How does Lindeberg CLT imply classicial CLT? For a sequence of zero-mean $\sigma^2$-variance IID random variables $X_1, \dots,$ the Linderberg condition is applied this way:
$\forall \epsilon >0$, as $n$ goes to $\infty$,
$$\sum_{i=1}^n E(\frac{|X_i|^2}{n\sigma^2} \times I_{|X_i| > \epsilon \sqrt{n}\sigma}) \to 0$$
which is by IID
$$E(\frac{|X_1|^2}{\sigma^2} \times I_{|X_1| > \epsilon \sqrt{n}\sigma}) \to 0$$
same as
$$E(|X_1|^2 \times I_{|X_1| > \epsilon \sqrt{n}}) \to 0$$
If I am all correct in the process, how is the last one true? Thanks in advance!

For every nonnegative random variable $Z$, the property that $E[Z\,\mathbf 1_{Z\geqslant z}]\to0$ when $z\to\infty$ is equivalent to $Z$ being integrable. (Proof: Lebesgue dominated convergence theorem.)
Edit: To prove that if $Z$ is integrable then $E[Z\,\mathbf 1_{Z\geqslant z}]\to0$, consider $Z_n=Z\,\mathbf 1_{Z\geqslant n}$. Note that $Z_n\to0$ almost surely and that $Z_n\leqslant Z$ with $Z$ integrable hence all the pieces to apply Lebesgue dominated convergence theorem are in place.
