I have been tasked with showing that for a metric space $(X,d)$, a subset $E \subseteq X$ is relatively compact $\iff$ $E$ is totally bounded. I believe I have shown the forward implication $(\Rightarrow)$. However, I'm struggling to show the backward implication.
For clarity, a set $E$ is said to be relatively compact if its closure $\overline{E}$ is compact.
Attempt at a proof:
Assume $E$ is totally bounded. Then by a corollary, we know that every sequence in $E$ has a Cauchy subsequence. Since $\overline{E}$ contains all limits points of $E$, we know that every Cauchy subsequence converges to some point in $\overline{E}$. Then $\overline{E}$ is sequentially compact (every sequence has a convergent subsequence). This implies that $\overline{E}$ is compact, and therefore $E$ is relatively compact.
My issue arises from the fact that I'm not considering what happens to sequences living inside of $\overline{E}\backslash E = \partial E$. I don't know if those converge or have Cauchy subsequences or anything. I only know things about sequences in $E$. Maybe I'm missing something obvious, or perhaps this entire proof is offbase. Any help would be appreciated.