Let $X$ denote a metric space.
Supposing every subset of $X$ that is closed and totally bounded is also compact, is $X$ necessarily complete?
What I've got so far. Assume every $X$ subset of $X$ that is closed and totally bounded is also compact, and consider a Cauchy sequence $a : \omega \rightarrow X.$ Then the image of $a$ (call it $A$) is totally bounded. (However, $A$ is not necessarily closed; now what?)
As you wrote, consider the image of the cauchy sequence, $A$, which is totally bounded by the cauchy property. So it's closure, $B$, is totally bounded and closed, so compact by hypothesis.
To finish the proof just use sequential compactness. You know a subsequence converges, but it is a basic fact (easy to prove) that if a subsequence of a cauchy sequence converges then the cauchy sequence converges.