Paired t-test means that the variances of the 2 samples are the same? Do paired samples imply that they have the same variance?
 A: Well consider the hypothesis for a paired T-test:
 

Where 
. In effect, you are actually taking the 

which is the variance of difference, and not the variance of the two groups. That is to say  
A: To illustrate @AndrewGustar's Comment, suppose you want to explore
the effect of a training method on the scores of subjects taking a particular
kind of exam.
Without intervention, the exam scores are $\mathsf{Norm}(\mu = 100, \sigma = 15).$ The intervention changes scores of individual subjects differently, according to $\mathsf{Norm}(10, 5).$ We can model this situation for
$n = 20$ subjects in R as follows:  
set.seed(520)
x      = rnorm(20, 100, 15)    # scores before
effect = rnorm(20, 8, 5)       # training effect
y      = x + effect            # scores after

For these data, a paired t-test detects the training effect, rejecting the null hypothesis that the population mean is unchanged; the p-value is quite small.
t.test(y, x, pair=T)

    Paired t-test

data:  y and x
t = 7.4702, df = 19, p-value = 4.566e-07
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 5.293207 9.413863
sample estimates:
mean of the differences 
               7.353535 

In the population, the original scores have a standard deviation of $15;$
the training introduces additional variability so that the standard deviation after training is $\sqrt{15^2 + 5^2} = 15.81.$ In the simulated exam scores, before and after scores reflect this increase:
sd(x); sd(y)
[1] 15.5792
[1] 16.14094

The paired t-test is precisely equivalent to a one-sample t test on the differences:
t.test(effect)

        One Sample t-test

data:  effect
t = 7.4702, df = 19, p-value = 4.566e-07
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 5.293207 9.413863
sample estimates:
mean of x 
 7.353535 

Note:
 In a real experiment one would probably not want to use exactly the
same exam before and after training, because that would confound the effect of training with the passage of time and familiarity with the exam.
One could use two equivalent versions of the exam, A and B. Half of the subjects could take A before training and the other half would take B before training. (Then a two-sample t test could be used to check whether the sequencing
of the versions matters.)
A: There's no need to assume the pair-members have the same variance as long as you work with the usual estimate of the standard error of the difference based off the standard deviation of the pair-differences.
[If you have some formula based on the correlation between them it might be possible there's been an (unnecessary) assumption made of equality, but I haven't seen it done that way.]
