Proving IID Central Limit Theorem using Lindeberg Conditions. The goal is to prove the IID Central Limit Theorem through Lindeberg's Condition.

Suppose that $X_1,X_2,\ldots\displaystyle\sim\text{i.i.d.}$ with $E[X_i]=\mu$ and $Var[X_i]=\sigma^2<\infty$.
Let $Y_i=X_i-\mu$ and $s_n^2=\sum_{i=1}^{n}Var[Y_i]=n\sigma^2$.
Prove that $Z_n:=\frac{\sum_{k=1}^{n}(X_k-\mu)}{s_n}\rightarrow$N$(0,1)$ in distribution.

Lindeberg's condition is as follows:

If the following holds:
$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{s_n^2}\sum_{i=1}^{n}E\big[Y_i^2\cdot\mathbb{I}_{( \ |Y_i|\geq\epsilon\cdot s_n \ )}\big]=0$ for all $\epsilon>0$
Then $Z_n\rightarrow$N$(0,1)$ in distribution.

So going right to lindbergs condition:
$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{n\sigma^2}\sum_{i=1}^{n}E\big[Y_i^2\cdot\mathbb{I}_{( \ |Y_i|\geq\epsilon\cdot \sigma\sqrt{n} \ )}\big]=\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\sigma^2}E\big[Y_i^2\cdot\mathbb{I}_{( \ |Y_i|\geq\epsilon\cdot \sigma\sqrt{n} \ )}\big]$
Now at this point I do know I can use Lebesgue Dominated Convergence theorem due to the following:
$ |Y_i^2\cdot\mathbb{I}_{( \ |Y_i|\geq\epsilon\cdot \sigma\sqrt{n} \ )}|=Y_i^2\cdot\mathbb{I}_{( \ |Y_i|\geq\epsilon\cdot \sigma\sqrt{n} \ )}\leq Y_i^2$ and $Y_i^2$ is integrable as $E[Y_i^2]=\sigma^2<\infty$
This means:
$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\sigma^2}E\big[Y_i^2\cdot\mathbb{I}_{( \ |Y_i|\geq\epsilon\cdot \sigma\sqrt{n} \ )}\big]=\frac{1}{\sigma^2}E\big[\displaystyle\lim_{n\rightarrow\infty}Y_i^2\cdot\mathbb{I}_{( \ |Y_i|\geq\epsilon\cdot \sigma\sqrt{n} \ )}\big]$

Now I do not understand how this above is zero, and that's where I am stuck.

I do know that by Markov's inequality:
$P(|Y_i|\geq\epsilon\sigma\sqrt{n})\leq\frac{E\big[|Y_i|\big]}{\epsilon\sigma\sqrt{n}}\rightarrow0$ as $n\rightarrow\infty$ as $E[|Y_i|]<\infty$, and so the probability of this event becomes zero.
Any help with understand this final step would be much appreciated!

All you need is $EY_1^{2} I_{\{|Y_1| >\epsilon \sigma \sqrt n\}} \to 0$ as $n \to \infty$ and this follows from DCT. [$Y_1^{2} I_{\{|Y_1| >\epsilon \sigma \sqrt n\}} $ is dominated by $Y_1^{2}$ which is integrable. Of course, the events $\{|Y_1| >\epsilon \sigma \sqrt n\}$ decrease to empty set so  $Y_1^{2} I_{\{|Y_1| >\epsilon \sigma \sqrt n\}} \to 0$ almost surely. ]. 
