The author of my intro analysis text has an exercise to give a proof of Bolzano-Weierstrass using axiom of completeness directly.
Let $(a_n)$ be a bounded sequence, and define the set $$S=\{x \in \mathbb{R} : x<a_n \text{ for infinitely many terms } a_n\}.$$
Show that there exists a subsequence $a_{n_k}$ converging to $s = \sup S$.
I feel I am close. I know that for any $\epsilon > 0$, there must be infinitely many $a_n$ such that $\sup S - \epsilon < a_n < \sup S + \epsilon$. (If there were only finitely many $a_n$ in that interval, then $\sup S + \frac{\epsilon}{2} \in S$, contradicting $\sup S$ as an upper bound.) However, I don't know how to pinpoint a single subsequence $(a_{n_k})$ such that all such elements with $k \geq \text{ some } N$ are in this interval for all $\epsilon$.