Let $n_{k}$ be such that $\sum_{k}E(|X_{n_{k}}|)<+\infty$,
In that case: $$E\left(\sum_{k}|X_{n_{k}}|\right)=\sum_{k}E(|X_{n_{k}}|)<+\infty$$ (monotone convergence)
Therefore, $\sum_{k}|X_{n_{k}}|<+\infty$ almost surely.
It follows that: $$\lim_{n\to+\infty}X_{n_{k}}=0$$ (as the general term of a convergent series).
How is monotone convergence used to demonstrate that, for a sequence $n_k$ such that $\sum_{k}E(|X_{n_{k}}|) < +\infty$, we can conclude that $E(\sum_{k}|X_{n_{k}}|) = \sum_{k}E(|X_{n_{k}}|)$, and how does this imply that the series $\sum_{k}|X_{n_{k}}|$ converges almost surely, leading to the conclusion $\lim_{n\to+\infty}X_{n_{k}} = 0$?
$\bullet$ Set $$ S_m = \sum_{k = 1}^m \vert X_{n_k}\vert $$ Then $S_m \le S_{m + 1} \ \forall m \in \mathbb{N}$ and $S_m \uparrow \sum_k \vert X_{n_k} \vert$. Now, using the monotone convergence theorem (or simply the Beppo Levi's theorem), we get $$ \mathbb{E}\left(\sum_k \vert X_{n_k} \vert\right) = \sum_k \mathbb{E}\vert X_{n_k} \vert $$
$\bullet$ If we further assume that $\mathbb{E}\left(\sum_k \vert X_{n_k} \vert\right) < \infty$. Then, $\sum_k \vert X_{n_k} \vert < \infty$ almost surely. Indeed, if $\mathbb{P}(\sum_k \vert X_{n_k} \vert = \infty) > 0$, then $$ \mathbb{E}\left(\sum_k \vert X_{n_k} \vert\right) \ge \mathbb{E}\left(\sum_k \vert X_{n_k} \vert \cdot 1\{\sum_k\vert X_{n_k} \vert = \infty\}\right) = \infty \cdot \mathbb{P}(\sum_k \vert X_{n_k} \vert = \infty) = \infty $$
$\bullet$ Finally, $\sum_k \vert X_{n_k} \vert < \infty$ almost surely implies $X_{n_k} \rightarrow 0$ almost surely. This fact is purely from real analysis. That is, $$ \sum_{n = 1}^\infty a_n < \infty \Longrightarrow \lim_n a_n = 0 $$