If $C(X)$ is the space of continuous functions on a compact space $X$, and $U$ is a subset of $C(X)$, then if $U$ is totally bounded then $U$ is also uniformly bounded and equicontinuous. In my textbook this is a lemma leading up to the proof of the Arzela-Ascoli Theorem.
Anyway, it's just interesting that this implication is only one way, because as soon as we take the closure of the totally bounded space we get that it is now compact and thus by the Azerla-Ascoli theorem it is closed, uniformly bounded, and equicontinuous. So adding closed-ness makes the implication go the other way also.
Yeah, if anyone could say a few words on what's going on here I would really appreciate it.
The converse is true as well, and this follows from either the proof or statement of Arzela-Ascoli, depending how you state it. Specifically, the key fact is that (the proof of) Arzela-Ascoli does not just characterize compact subsets of $C(X)$, but also subsets whose closures are compact: namely, a set $U\subseteq C(X)$ has compact closure iff it is uniformly bounded and equicontinuous. In particular, if $U$ is uniformly bounded and equicontinuous, then $\overline{U}$ is compact, so $\overline{U}$ is totally bounded. Since $U\subseteq\overline{U}$, this implies $U$ is also totally bounded.
(Alternatively, if your version of Arzela-Ascoli only characterizes compact sets and not just sets whose closures are compact, just note that uniform boundedness and equicontinuity are preserved by taking closures (checking this is just some straightforward $\epsilon$-$\delta$ arguments). So then your version of Arzela-Ascoli applies to $\overline{U}$ to show it is compact.)