Does a "Normal Distribution" need to have mean=median=mode? I've been in a debate with my graduate-level statistics professor about "normal distributions".  I contend that to truly get a normal distribution one must have mean=median=mode, all the data must be contained under the bell curve, and perfectly symmetrical around the mean.  Therefore, technically, there are virtually NO normal distributions in real studies, and we should call them something else, perhaps "near-normal".  
She says I'm too picky, and if the skew/kurtosis are less than 1.0 it is a normal distribution and took off points on an exam. The dataset is total number of falls/year in a random sampling of 52 nursing homes which is a random sample of a larger population.   Any insight?
Problem:

QUESTION: 3.  Compute measures of skewness and kurtosis for this data. Include a histogram with a normal curve. Discuss your findings. Is the data normally distributed? 
Statistics 
Number of falls  
N  Valid    52
   Missing   0
Mean        11.23
Median      11.50
Mode         4a

a. Multiple modes exist. The smallest value is shown
Number of falls  
N  Valid    52
   Missing   0
Skewness      .114
Std. Error of Skewness    .330
Kurtosis  -.961
Std. Error of Kurtosis    .650


My answer:

The data is platykurtic and has only slight positive skewing, and it is NOT a normal distribution because the mean and median and mode are not equal and the data is not evenly distributed around the mean.  In reality virtually no data is ever a perfect normal distribution, although we can discuss “approximately normal distributions” such as height, weight, temperature, or length of adult ring finger in large population groups.

Professor's answer:

You are correct that there is no perfectly normal distribution.  But, we are not looking for perfection.  We need to look at data in addition to the histogram and the measures of central tendency.  What do the skewness and kurtosis statistics tell you about the distribution? Because they are both between the critical values of -1 and +1, this data is considered to be normally distributed.

 A: The teacher is clearly out of his/her element, and probably should not be teaching statistics. It seems worse to me to teach something wrong than to not teach it at all. 
These issues could all be cleared up easily if the distinction between "data" and "process that produced the data" were made more clearly.  Data target the process that produced the data. The normal distribution is a model for this process.  
It makes no sense to talk about whether the data are normally distributed. For one reason, the data are always discrete. For another reason, the normal distribution describes an infinity of potentially observable quantities, not a finite set of specific observed quantities.
Further, the answer to the question "is the process that produced the data a normally distributed process" is also always "no," regardless of the data. Two simple reasons: (i) any measurements we take are necessarily discrete, being rounded off to some level. (ii) perfect symmetry, like a perfect circle, does not exist in observable nature. There are always imperfections.  
At best, the answer to the question "what do these data tell you about normality of the data-generating process" could be given as follows: "these data are consistent with what we would expect to see, had the data truly come from a normally distributed process."  That answer correctly does not conclude that the distribution is normal.
These issues are very easily understood by using simulation. Just simulate data from a normal distribution and compare those to the existing data.  If the data are counts (0,1,2,3,...), then obviously the normal model is wrong because it does not produce numbers like 0,1,2,3,...; instead, it produces numbers with decimals that go on forever (or at least as far as the computer will allow.)  Such simulation should be the first thing you do when learning about the normality question. Then you can more correctly interpret the graphs and summary statistics.
A: You're missing the point and probably are also being "difficult," which is not appreciated in the industry. She's showing you a toy example, to train you in assessment of normality of a data set, which is to say whether the data set comes from a normal distribution. Looking at distribution moments is one way to check the normality, e.g. Jarque Bera test is based on such an assessment.
Yes, the normal distribution is perfectly symmetrical. However, if you draw a sample from a true normal distribution, that sample will most likely not be perfectly symmetrical. This is the point you're completely missing. You can test this very easily yourself. Just generate a sample from Gaussian distribution, and check its moment. They'll never be perfectly "normal," despite the true distribution being such.
Here's a silly Python example. I'm generating 100 samples of 100 random numbers, then obtaining their means and medians. I print the first sample to show that the mean and median are different, then show the histogram of the difference between the means and medians. You can see that it's rather narrow, but the difference is basically never zero. Note, that the numbers are truly coming from a normal distribution. 
code:
import numpy as np
import matplotlib.pyplot as plt

np.random.seed(1)
s = np.random.normal(0, 1, (100,100))
print('sample 0 mean:',np.mean(s[:,0]),'median:',np.median(s[:,0]))

plt.hist(np.mean(s,0)-np.median(s,0))
plt.show()
print('avg mean-median:',np.mean(np.mean(s,0)-np.median(s,0)))

outputs:

P.S.
Now, whether the example from your question should be considered normal or not depends on the context. In the context of what was taught in your class room you're wrong, because your professor wanted to see whether you know the rule of thumb test that she gave you, which is that skew and excess kurtosis need to be in -1 to 1 range. 
I personally never used this particular rule of thumb (I can't call it a test), and didn't even know it existed. Apparently, some people in some fields do use it though. If you were to plug your data set descriptives into JB test, it would have rejected normality. Hence, you're not wrong to suggest that the data set is not normal, of course, but you're wrong in a sense that you failed to apply the rule that was expected from you based on what's been taught in the class. 
If I were you I'd politely approach your professor and explain myself, as well as show JB test output. I'd acknowledge that based on her test my answer was wrong, of course. If you attempt to argue with her the way you argue here, your chances are very low to get the point back in the test, because your reasoning is weak about medians and means and samples, it shows lack of understanding of samples vs. populations. If you change your tune, then you'll have a case.
A: I'm an engineer, so in my world, the applied statistician is what I see most, and get the most concrete value.  If you are going to work in applied, then you need to be solidly grounded in practice over theory: whether or not it is elegant, the aircraft has to fly and not crash.    
When I think about this question the way I approach it, as many of my technical betters here have also done, is to think about "what does it look like in the real world with the presence of noise".
The second thing that I do is, often, to make a simulation that allows me to get my hands around the question.
Here is a very brief exploration:
#show how the mean and the median  differ with respect to sample size

#libraries
library(reshape2)
library(ggplot2)

#sample sizes
ssizes <- 10^(seq(from=1, to=3, by=0.25))
ssizes <- round(ssizes)

#loops per sample
n_loops <- 5000

#pre-declare, prep for loop
my_store <- matrix(0, 
                   ncol = 3, 
                   nrow = n_loops*length(ssizes))

count <- 1

for(i in 1:length(ssizes)){

  #how many samples
  n_samp <- ssizes[i]

  for(j in 1:n_loops){

    #draw samples
    y <- 0
    y <- rnorm(n = n_samp,mean = 0, sd = 1)

    #compute mean, median, mode
    my_store[count,1] <- n_samp
    my_store[count,2] <- median(y)
    my_store[count,3] <- mean(y)


    #update
    count = count + 1
  }
}


#make data into ggplot friendly form
df <- data.frame(my_store)
names(df) <- c("n_samp", "median","mean")

df <- melt(df, id.vars = 1, measure.vars = c("median","mean"))


#make ggplot
ggplot(df, aes(x=as.factor(n_samp), 
               y = value, 
               fill = variable)) + geom_boxplot() + 
  labs(title = "Contrast Median and Mean estimate variation vs. Sample Size",
       x = "Number of Samples",
       y = "Estimated value")

It gives this as the output:

Note: be careful about the x-axis, because it is log-scaled, not uniform-scaled.
I know that the mean and median are exactly the same.  The code says it.  The empirical realization is greatly sensitive to sample size, and if there aren't truly infinite samples, then they can't ever perfectly match with theory.
You can think about whether the uncertainty in the median envelopes the estimated mean or vice versa.  If the best estimate of the mean is within the 95% CI of the estimate for the median, then the data can't tell the difference.  The data says they are the same in theory.  If you get more data, then see what it says.  
A: In medical statistics, we only ever comment on the shapes and seeming of distributions. The fact that no discrete finite sample can ever be normal is irrelevant and pedantic. I would mark you wrong for that.
If a distribution looks "mostly" normal, we are comfortable with calling it normal. When I describe distributions for a non-statistical audience, I am very comfortable with calling something approximately normal even when I know the normal distribution is not the underlying probability model, I get the sense I would side with your teacher here... but we have no histogram or dataset to verify.
As a tip, I would go through the following inspections very closely: 


*

*who are the outliers, how many and what are their values? 

*Are the data bimodal? 

*Do the data seem to take a skewed shape so that some transformation (like a log) would better quantify the "distance" between observations?

*Is there apparent truncation or heaping so that assays or labs are failing to reliably detect a certain range of values?

A: A problem with your discussion with the professor is one of terminology, there's a misunderstanding that is getting in the way of conveying a potentially useful idea. In different places, you both make errors.
So the first thing to address: it's important to be pretty clear about what a distribution is.
A normal distribution is a specific mathematical object, which you could consider as a model for a process (which you might consider an uncountably infinite population of values; no finite population can actually have a continuous distribution).
Loosely, what this distribution does (once you specify the parameters) is define (via an algebraic expression) the proportion of the population values that lies within any given interval on the real line. Slightly less loosely, it defines the probability that a single value from that population will lie in any given interval.
An observed sample doesn't really have a normal distribution; a sample might (potentially) be drawn from a normal distribution, if one were to exist. If you look at the empirical cdf of the sample, it's discrete. If you bin it (as in a histogram) the sample has a "frequency distribution", but those aren't normal distributions. The distribution can tell us some things (in a probabilistic sense) about a random sample from the population, and a sample may also tell us some things about the population.
A reasonable interpretation of a phrase like "normally distributed sample"* is "a random sample from a normally distributed population".
*(I generally try to avoid saying it myself, for reasons that are hopefully made clear enough here; usually I manage to confine myself to the second kind of expression.)
Having defined terms (if still a little loosely), let us now look at the question in detail. I'll be addressing specific pieces of the question.

normal distribution one must have mean=median=mode

This is certainly a condition on the normal probability distribution, though not a requirement on a sample drawn from a normal distribution; samples may be asymmetric, may have mean differ from median and so on. [We can, however, get an idea how far apart we might reasonably expect them to be if the sample really came from a normal population.]

all the data must be contained under the bell curve

I am not sure what "contained under" means in this sense.

and perfectly symmetrical around the mean.

No; you're talking about the data here, and a sample from a (definitely symmetrical) normal population would not itself be perfectly symmetric.
Here's some simulated samples from normal distributions:

If you generate a number of samples of about that sample size (60) and plot histograms with about 10 bins, you may see similar variation in general shape.
As you can see from the histograms, these are not actually symmetric. Some, like 2, 4 and 7, are quite distinctly asymmetrical. Some have quite short tails, like 5 and 8, some have noticeably longer tails, at least on one side. Some suggest multiple modes. None actually look all that close to what an actual normal density looks like $-$ that is, even random samples don't necessarily look all that much like their populations, at least not until the sample sizes are fairly large $-$ considerably larger than the n=60 I used here.

Therefore, technically, there are virtually NO normal distributions in real studies,

I agree with your conclusion but the reasoning is not correct; it's not a consequence of the fact that data are not perfectly symmetric (etc); it's the fact that populations are themselves not perfectly normal.

if the skew/kurtosis are less than 1.0 it is a normal distribution

If she said this in just that way, she's definitely wrong.
A sample skewness may be much closer to 0 than that (taking "less than" to mean in absolute magnitude not actual value), and the sample excess kurtosis may also be much closer to 0 than that (they might even, whether by chance or construction, potentially be almost exactly zero), and yet the distribution from which the sample was drawn might be distinctly non-normal (e.g. bimodal, or clearly asymmetric, or perhaps with somewhat heavier tails than the normal $-$ it's not just the tail that determines kurtosis)
We can go further -- even if we were to magically know the population skewness and kurtosis were exactly that of a normal, it still wouldn't of itself tell us the population was normal, nor even something close to normal.
Here's an example:

This particular example is strongly bimodal, heavier tailed than the normal, but symmetric. It has the same skewness and kurtosis as the normal.
Further examples can be found in this answer.
Not all are symmetric, and some are discrete.

The dataset is total number of falls/year in a random sampling of 52 nursing homes which is a random sample of a larger population.

The population distribution of counts are never normal. Counts are discrete and non-negative, normal distributions are continuous and over the entire real line.
But we're really focused on the wrong issue here. Probability models are just that, models. Let us not confuse our models with the real thing.
The issue isn't "are the data themselves normal?" (they can't be), nor even "is the population from which the data were drawn normal?" (this is almost never going to be the case).
A more useful question to discuss is "how badly would my inference be impacted if I treated the population as normally distributed?"
That is we should not be overly focused on whether the assumption is true (we shouldn't expect that), but whether it's useful, or perhaps what and how severe might  the consequences be if we were to use such a model.
It's also a much harder question to answer well, and may require considerably more work than glancing at a few simple diagnostics.
The sample statistics you showed are not particularly inconsistent with normality (you could see statistics like that or "worse" not terribly rarely if you had random samples of that size from normal populations), but that doesn't of itself mean that the actual population from which the sample was drawn is automatically "close enough" to normal for some particular purpose. It would be important to consider the purpose (what questions you're answering), and the robustness of the methods employed for it, and even then we may still not be sure that it's "good enough"; sometimes it may be better to simply not assume what we don't have good reason to assume a priori (e.g. on the basis of experience with similar data sets).

it is NOT a normal distribution

Data - even data drawn from a normal population - never have exactly the properties of the population; from those numbers alone you don't have a good basis to conclude that the population is not normal here.
On the other hand neither do we have any reasonably solid basis to say that it's "sufficiently close" to normal - we haven't even considered the purpose of assuming normality, so we don't know what distributional features it might be sensitive to.
For example, if I had two samples for a measurement that was bounded, that I knew would not be heavily discrete (not mostly only taking a few distinct values) and reasonably near to symmetric, I might be relatively happy to use a two-sample t-test at some not-so-small sample size; it's moderately robust to mild deviations from the assumptions (somewhat level-robust, somewhat less power-robust). But I would be considerably more cautious about as causally assuming normality when testing equality of spread, for example, because the best test under that assumption is quite sensitive to the assumption.

Because they are both between the critical values of -1 and +1, this data is considered to be normally distributed."

If that's really the criterion by which one decides to use a normal distributional model, then it will sometimes lead you into quite poor analyses.
The values of those statistics do give us some clues about the population from which the sample was drawn, but that's not at all the same thing as suggesting that their values are in any way a 'safe guide' to choosing an analysis.
Cast your mind back to the fact that there's distributional examples where the population has very different shape from the normal, but with the same population skewness and kurtosis. Add to that the inherent noise in their sample equivalents (and not least of all, the considerable downbias typical of sample kurtosis), and you may well be concluding rather too much on very limited and possibly misleading evidence.

Now to address the underlying issue with even a better phrased version of such a question as the one you had:
The whole process of looking at a sample to choose a model is fraught with problems -- doing so alters the properties of any subsequent choices of analysis based on what you saw! e.g for a hypothesis test, your significance levels, p-values and power are all not what you would choose/calculate them to be, because those calculations are predicated on the analysis not being based on the data.
See, for example Gelman and Loken (2014), "The Statistical Crisis in Science," American Scientist,  Volume 102, Number 6, p 460
(DOI: 10.1511/2014.111.460) which discusses issues with such data-dependent analysis.
A: I think you and your professor are talking in different context. Equality of mean = median = mode is characteristics of theoretical distribution and this is not the only characteristics. You can not say that if for any distribution above property hold then distribution is normal. T-distribution is also symmetric but it is not normal. So, you are talking about theoretical properties of normal distribution which hold always true for normal distribution. 
You professor is talking about distribution of sample data. He is right, you will never get data in real life, where you will find mean = median = mode. This is simply due to sampling error. Similarly, it is very unlikely, you will get zero coefficient of skewness for sample data and zero excess kurtosis. Your professor is just giving you simple rule to get an idea about the distribution from the sample statistics. Which is not true in general (without getting further information). 
A: For practical purposes, underlying processes such as this one are usually finely approximated by normal distribution without anyone raising an eyebrow.
However, if you wanted to be pedantic the underlying process in this case can't be normally distributed, because it can't produce negative values (number of falls can't be negative). I wouldn't be surprised if it was in fact at least a bi-modal distribution with second peak close to zero.
