Let S be totally bounded. So, $\forall\epsilon>0$, $S$ can be covered by a finite number of balls of radius $\epsilon$. Now, let $\{S_n\}$ be a cover of $S$ with open sets. So, $S_n\cap S$ is also totally bounded by say set $U_n$ as subsets of totally bounded set is totally bounded. So, $S$ can be covered by $\cup U_n$ and as $S$ is totally bounded, we can have a finite subset of $\cup U_n$, say $U$, which covers $S$. Now, for each element of $U$, we can choose a corresponding set in $\{S_n\}$ to which it belongs. Thus, we can get a finite subset of $\{S_n\}$ that covers S.
But here I didn't require $S$ to be closed/complete. Is it necessary or does being "totally bounded" takes care of being "closed or complete"? Can you please tell where I went wrong.
The open interval $(0,1)$ (with the standard metric) is totally bounded but not compact (for example the intervals $(1/n,1)$ are an open cover with no finite subcover). The problem with the proof is that there's simply no reason each element of $U$ should be contained in some $S_n$.
In the proof that totally bounded plus complete implies compact we first show that any open cover has a "Lebesgue number". Which means precisely that if $(S_n)$ is an open cover and $U$ is the cover by balls of radius $\delta$, then if $\delta$ is small enough every element of $U$ is contained in some $S_n$.