CLT - infinite variance Student t distribution with df=2 has infinite variance => CLT should not hold(?)
Yet, when I draw a large sample thousands of times and plot the histogram of the standardized sample mean, it has a normal distribution. Why is this? 
 A: The distribution in question has a form given in this Wiki article. Note that for 1 degree of freedom, the t-distribution becomes a Cauchy distribution. The Cauchy is a member of the class of "stable" distributions, hence, the average of two cauchy distributions is again cauchy. The same applies for 2 df (although it doesn't have a nice name).
What does this mean? It means that you are standardizing distributions with infinite kurtosis. Large kurtosis distributions present practical challenges when you are simulating them, due to their ability to produce "black swans" (to steal from Taleb)...which are values that are orders of magnitude above a typical value, such that they effectively "wash out" all values that came before. These extreme outliers are quite rare; unfortunately, these outliers cannot be dismissed as inconsequential (as can be done with a normal distribution), since when they occur, they are not just "high values" but "game changers".
By taking a large sample and normalizing, you are creating a distribution that, for all intents and purposes, appears very normal for the central 99%...however, its that pesky 1% in the tails that will undo any unfortunate analyst who concludes that the risks are normally distributed, and hence well-behaved...2008 anyone?
Technical note: most "goodness of fit" methods are not very sensitive to the tails of the dataset, hence you are unlikely to detect such an effect with such a test. 
A: This seems to be an old question, but I find it interesting hence a treatment: 
We want to show that $\displaystyle X_n=\frac{\sum x_i}{s_n} \sim N(0,1)$ for an appropriate choice of $\displaystyle s_n \rightarrow \infty $. To do so recall the Lindeberg's condition that $\displaystyle lim_{n \rightarrow \infty} \sum \int_{|z|\geq s_n a}\frac{z^2}{s_n^2}dF_i(z)=0$ for all $ \displaystyle a>0$ is equivalent to $\displaystyle X_n \sim N(0,1)$ and $\displaystyle \lim_{n \rightarrow \infty} \max_{1\leq i \leq n}\Pr{\frac{|x_i|}{s_n}\geq a}=0$. 
Using the "characteristic function approach", $ \displaystyle \phi(t)$, our task becomes showing $ \displaystyle \lim_{n \rightarrow \infty}n \ln\phi(t/s_n)=-t^2/2$. 
First note that for an iid sequence we clearly have that $\displaystyle \lim_{n \rightarrow \infty} \max_{1\leq i \leq n}\Pr{\frac{|x_i|}{s_n}\geq a}=0$ thus $\displaystyle \lim_{n \rightarrow \infty}|\phi(t/s_n)-1| =0$, therefore 
$\begin{align}
\lim_{n \rightarrow \infty}n [\phi(t/s_n)-1]&=-\frac{t^2}{2}\lim_{n \rightarrow \infty}n \int_{|z|<s_n a}\frac{z^2}{s_n^2}\frac{1}{(2+z^2)^{3/2}}dz\\
&=-\frac{t^2}{2}\lim_{n \rightarrow \infty}\frac{n}{s_n^2} \Big(2 \mathrm{arcsinh}(s_n a/\sqrt{2})-2\frac{s_n a}{\sqrt{1+s_n^2 a^2}}\Big)\\
&=-\frac{t^2}{2}\lim_{n \rightarrow \infty}\frac{2 n \log{s_n}}{s_n^2} 
\end{align}$
Now choose $\displaystyle s_n=\sqrt{n \log n}$ to have $ \displaystyle \lim_{n \rightarrow \infty}\frac{2 n \log{s_n}}{s_n^2}=1$. 
Therefore we need $\displaystyle s_n \sim n \log n$, a rate a bit faster than $\displaystyle n$ which we need when the variance is bounded.
