Let $X:\Omega\times[0,\infty)\to\mathbb R$ be a right-continuous uniformly integrable martingale on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0}\operatorname P)$ and $(t_n)_{n\in\mathbb N}\subseteq[0,\infty)$ be nondecreasing with $t_n\xrightarrow{n\to\infty}\infty$.
Why can we find an $X_\infty\in L^1(\operatorname P)$ with $X_{t_n}\xrightarrow{n\to\infty}X_\infty$ almost surely? And why does this $X_\infty$ not depend on $(t_n)_{n\in\mathbb N}$?
Clearly, since $X$ is uniformly integrable, $(X_{t_n})_{n\in\mathbb N}$ is $L^1$-bounded and hence there is some increasing $(n_k)_{k\in\mathbb N}\subseteq\mathbb N$ with $n_k\to\infty$ and some $\tilde X_\infty$ with $\left\|\tilde X_\infty-X_{t_{n_k}}\right\|_{L^1}\xrightarrow{k\to\infty}0$. But that doesn't yield the claim.
Note that $X_{t_n}$ is an $L^1(P)$-bounded discrete time martingale with respect to the filtration $(\mathcal{F}_{t_n})_{n \in \mathbb{N}}$. So by the discrete time Martingale Convergence Theorem there is $X_{\infty} \in L^1(P)$ such that $X_{t_n} \to X_{\infty}$ almost surely. (And in fact since $(X_{t_n})_{n \in \mathbb{N}}$ is uniformly integrable this limit is also in $L^1$-norm)
We are left to check that this limit doesn't depend on choice of the sequence $(t_n)$. Given an alternative such sequence $(s_n)$ form the sequence $(\tilde{t}_n)$ by putting the set $\{s_n \mid n \in \mathbb{N} \} \cup \{t_n \mid n \in \mathbb{N} \}$ in nondecreasing order.
$(\tilde{t}_n)$ has the same properties as $(t_n)$ so as above, $X_{\tilde{t}_n} \to \tilde{X}_\infty$ almost surely for some $\tilde{X}_\infty$. But since the subsequence $X_{t_{n}} \to X_\infty$ almost surely we have $\tilde{X}_\infty = X_\infty$ almost surely.