Does Student's T require normally-distributed data? Student's T distribution can be derived from the fact that:
$$ \frac{\frac{\bar X - \mu}{\sigma/\sqrt{n}}}{\sqrt{S^2/\sigma^2}} $$
Is an approximately normally distributed variable in the numerator (central limit theorem) with (the square root of) a chi-squared random variable in the denominator.  However, does the chi-square result not require that $X$ be normal?  If not, what is the reasoning here that allows us to use a chi-square distribution for the denominator?

It seems from the comments that the short answer here is for large samples, "no", because the sampling distribution of the mean converges to a normal distribution for large $n$.  For small samples, the answer is "maybe" depending on lots of things.  It seems from some simulations that one of the biggest factors is the skewness of your population distribution:
All of these are done taking 10000 samples of size n=10.  The "theoretical" sampling distributions are normal, chi-square and student's-t respectively.
Uniformly distributed population:

Beta distributed population with $\alpha = \beta = 0.5$

Weibull distributed population with $\lambda = 1, k = 0.5$

As $k$ gets smaller, the t distribution does a worse and worse job for the Weibull distributed population.
