I think what would help here is a good, solid overview of distributions and standard deviations / standard errors. I take it you know what a distribution is, but note that there are typically 3 different entities that we work with, all of which have distributions that are relevant.
First there is the population distribution, which is understood to be infinite. Usually, we think of there being an infinite number of units which could be observed; in your case, I think it's being conceived as one true value of the boiling point, but the measurement system has random error associated with it. That is, I don't think they have sampling error in mind, but either way, there are infinite potential values of the boiling point, and these values are distributed in some way. (I gather it's being assumed that the distribution is normal, or normal-ish--which is good enough for our purposes.)
The second is your sample distribution. You have the observations for your sample, you can describe this completely.
But how can you connect what you know about your sample to what you want to know about the population? This is where the third distribution comes in. It is the sampling distribution of the sample mean. Specifically, it is the distribution that you would have, if you followed the same procedure as here (i.e., taking 6 samples in the same way, measuring the boiling point in the same way, calculating the mean in the same way, etc.), over and over again infinitely. It is the sampling distribution that helps you draw inferences about the population (more precisely, the true boiling point of the liquid). This issue is the crux of your confusion. When they write, "then the sample mean will have the distribution", the problem is that it is not clearly worded. Your sample mean is a single realized value; it does not have a distribution, except in a trivial sense. Instead, they are referring to the sampling distribution of the sample mean in the sense I just described.
An important issue concerns the shape of the sampling distribution. The Central Limit Theorem (also discussed here) is often invoked to assert that the (hypothetical, infinite set of such) sample means would be distributed normally. Typically, I would feel much more comfortable saying that if the $N$ were considerably greater than 6(!), but the situation described is clearly meant to be understood as the distribution of the data are either normal or very close to, and therefore we may be OK here. The CLT doesn't only tell us that the sampling distribution will be normally distributed, it also tells us (quite usefully) that the standard deviation of the sampling distribution (called the 'standard error') is $\sigma/\sqrt{N}$. You can then use this standard error to calculate your 95% confidence interval. The CLT is really very helpful.
Finally, you are right that the population formula for the variance:
$$
Var(X)=\frac{\sum(x_i-\bar{x})^2}{N},
$$
when applied to a sample, provides a biased estimate of the true population variance. This is because you first used the very same sample data to calculate the mean, $\bar{x}$, and you need to account for this fact. This is done by dividing by $N-1$ instead of $N$. After having done this, however, you have appropriately accounted for that source of bias--you're done with that. When it comes to estimating the variability in the sampling distribution of the mean, you don't have to worry about it any more. Thus, when the CLT tells us that the standard error will be $\sigma/\sqrt{N}$, it is talking about a different thing, and is assuming that we have already taken care of the issue and that the $\sigma$ we are using is correct.
