Need to show if $X_n\to X$, in $\mathcal{L}^1,$ then some sequence $X_{nk}\to X$ a.s.
Let $\delta=2^{-k}$, then \begin{align*} \mathbb{E}^\mathbb{P}(|X_n-X|<\delta) &=\int |X_n-X| \, d\mathbb{P} \\ &= \int_{|X_n-X| \leq \epsilon} |X_n-X|\, d\mathbb{P} + \int_{|X_n-X| > \epsilon} |X_n-X| \, d\mathbb{P}\\ &\le \int_0^\infty \delta dk \end{align*} And so the limit of this value is $0$ as $n\to \infty$, meaning that it converges a.s.
I'm not too sure this is correct, any validation?
I can't really follow what you're trying to do there. It is definitely not true that $X_n$ itself converges a.s. It can even converge nowhere. So any direct argument like this cannot work.
The standard procedure for this (which, by the way, is a standard lemma in the proof of the Riesz-Fischer theorem) is to first extract a rapidly Cauchy subsequence and then use the Borel-Cantelli lemma. Specifically this means that you should obtain a subequence $X_{n_k}$ such that $\| X_{n_{k+1}} - X_{n_k} \|^{1/2}_{L^1}$ is a summable sequence of numbers.
With such a subsequence in hand, by using Markov's inequality you get $\mathbb{P}(|X_{n_{k+1}}-X_{n_k}| \geq \| X_{n_{k+1}}-X_{n_k} \|^{1/2}_{L^1}) \leq \| X_{n_{k+1}}-X_{n_k} \|_{L^1}^{1/2}$ which is summable, so by the Borel-Cantelli lemma, with probability $1$ one is in only finitely many of these sets. In other words you can discard finitely many $k$ to get a new subsequence such that $\mathbb{P}(|X_{n_{k+1}}-X_{n_k}| > \| X_{n_{k+1}}-X_{n_k} \|_{L^1}^{1/2})=0$ for all $k$. Now you're in business to argue that $X_{n_k}$ is Cauchy a.s. and thus convergent a.s., by looking far enough out in the sequence $X_{n_k}$ that $\sum_{k=K}^\infty \| X_{n_{k+1}}-X_{n_k} \|^{1/2}_{L^1} < \varepsilon$ for some $K=K(\varepsilon)$.