Which metric spaces are totally bounded?

A subset $S$ of a metric space $X$ is totally bounded if for any $r>0$, $S$ can be covered by a finite number of $X$-balls of radius $r$.

A metric space $X$ is totally bounded if it is a totally bounded subset of itself.

For example, bounded subsets of $\mathbb{R}^n$ are totally bounded.

Are there any interesting necessary and/or sufficient conditions for a metric space or its subsets to be totally bounded?

[Background: I was trying to generalize problem 4.8 of baby Rudin which asks you to prove that a real uniformly continuous function on a bounded subset $E$ of the real line is bounded. It seems after a little googling that a more general true statement would require $E$ to be a totally bounded subset of some metric space. But where might we meet such subsets?]

A metric space is totally bounded if and only if every sequence has a Cauchy subsequence.

(Try and prove this!)

As you might suspect, this is basically equivalent to what Jonas has said. The key between these two is provided by:

A metric space is compact if and only if it is totally bounded and complete.

In other words, every sequence has a convergent subsequence (compact) if and only if every sequence has a Cauchy sequence (Totally bounded) and every Cauchy sequence converges (complete).

A metric space is totally bounded if and only if its completion is compact. (However, a little googling reveals that totally bounded doesn't necessarily imply compact completion without the axiom of choice.) This gives one explanation of the result in your background, because uniformly continuous functions extend to completions.

For subsets of a complete metric space, total boundedness is equivalent to having compact closure, which is what we see for bounded subsets of $\mathbb{R}^n$.

I don't know, is this interesting?