I know this has been proved in other links, but I am wondering about the validity of the following proof:
Suppose $X_n$ is a Cauchy sequence in $L^\infty$. Then there exists a subsequence $Y_k \equiv X_{n_k}$ such that $\|Y_{k+1} - Y_k \|_\infty < \frac{1}{2^k}$.
Define $H_n \equiv \sum_{k=0}^{n-1}|Y_{k+1} - Y_k|$ and note that $\|H_n\|_\infty \leq \sum_{k=0}^\infty \|Y_{k+1} - Y_k\|_\infty \leq 2$ so that $H \equiv \lim_{n \rightarrow \infty} H_n$ is well defined and finite almost surely (so $H_n$ is Cauchy in $\mathbb{R}$ almost surely). Thus $Y_k$ is a Cauchy sequence in $\mathbb{R}$ since (WLOG put $m \ge n$)
$$|Y_m - Y_n| \leq H_m - H_n < \epsilon \quad \text{ for } m, n \text{ large}$$
and thus $Y_k$ is convergent to a limit, say $X$. To see that this is the $L^\infty$ limit for the subsequence, we note that $$\|Y_n - X\|_\infty \leq \|H- H_n\|_\infty \leq \sum_{k=n}^\infty\frac{1}{2^k} \xrightarrow{n \rightarrow \infty} 0$$
Now we note that there exists $N$ such that $m,n \ge N$ implies $\|X_m - X_n\|_\infty < \epsilon/2$ and such that $k \ge N$ implies $\|Y_k - X\|_\infty < \epsilon/2$ Thus for $n \ge N$, we obtain (choosing any $k \ge N$ and noting $n_k \ge k$),
$$\|X_n - X \|_\infty \leq \|X_n - X_{n_k} \|_\infty + \|X_{n_k} - X\|_\infty < \epsilon$$
and we are done. Does this work? Any feedback is much appreciated!