I know that there are other posts discussing proofs of the BW Theorem. However, I could not find a post that discussed my specific proof. So, I have made a new post.
If the proof is correct, but lacks refinement, please allow me your feedback. It will be greatly appreciated, as I'm not only trying to improve my understanding of math, but also my communication of it.
Let $(a_k)$ be a bounded sequence, and define the set $$S = \{x \in \mathbb{R}: x < a_k \textrm{ for infinitely many $a_k$}\}.$$
Given that $S$ is bounded, as $(a_k)$ is bounded, the set $S$ has a supremum. I will show that there exists a subsequence $(a_{k_i})$ of $(a_k)$ that converges to $\operatorname{sup}S,$ hence demonstrating that there exists a convergent subsequence of $(a_k).$
Note that for all $n \in \mathbb{N},$ there may be only finitely many elements $a_{k}$ in $(a_k)$ that satisfy $$\operatorname{sup}S + \frac{1}{n} \leq a_{k}.$$ If there were some $n \in \mathbb{N}$ for which the aforementioned inequality holds for infinitely many elements $a_{k}$ in $(a_k),$ the set $S$ would contain elements $s \in [\operatorname{sup}S, \operatorname{sup}S+\frac{1}{n}),$ contradicting that $\operatorname{sup}S$ is the supremum of $S.$
Also, note that for all $n \in \mathbb{N},$ there exists $s \in S$ such that $$\operatorname{sup}S - \frac{1}{n} < s < \operatorname{sup}S,$$ which implies that for all $n \in \mathbb{N},$ there exist infinitely many elements $a_k$ in $(a_k)$ that satisfy $$\operatorname{sup}S - \frac{1}{n} < s < a_k.$$
Given that for $n \in \mathbb{N}$ there exists only finitely many elements $a_k$ that satisfy $$\operatorname{sup}S + \frac{1}{n} \leq a_k,$$ it follows that there for all $n \in \mathbb{N}$ there exist infinitely many elements $a_k$ that satisfy $$\operatorname{sup}S - \frac{1}{n} < a_k \leq \operatorname{sup}S + \frac{1}{n}.$$
A subsequence $(a_{k_i})$ that converges to $\operatorname{sup}S$ has now become clear. Given $i \in \mathbb{N},$ there exists an element $a_{k_i}$ such that $$\operatorname{sup}S - \frac{1}{i} < a_{k_i} \leq \operatorname{sup}S + \frac{1}{i}.$$ As $k_i$ is finite, there are only finitely many elements that appear before $a_{k_i}$ in the sequence $(a_k).$ As there are infinitely many terms in the sequence $(a_k)$ that satisfy $$\operatorname{sup}S - \frac{1}{i+1} < a_k \leq \operatorname{sup}S + \frac{1}{i+1},$$ there exists $a_{k_{i+1}}$ such that $$\operatorname{sup}S - \frac{1}{i+1} < a_{k_{i+1}} \leq \operatorname{sup}S + \frac{1}{i+1} \quad \textrm{and} \quad k_{i+1} > k_{i}.$$
Hence, a convergent subsequence $(a_{k_i})$ may be constructed by choosing inductively terms in the sequence $(a_k)$ that satisfy $$\operatorname{sup}S - \frac{1}{i} < a_{k_i} \leq \operatorname{sup}S + \frac{1}{i}$$ for $i = 1, 2, ...$