Countable choice and totally bounded metric spaces

Question

Can we prove that the following statement is equivalent to the axiom of countable choice (CC)?

If every sequence in a metric space $X$ has a Cauchy subsequence, then $X$ is totally bounded.

Note: CC is known to be equivalent to the above condition with "metric space" replaced by "pseudometric space".

Answer 1

I found a proof in case we work with pseudometric space instead of metric space. If CC fails then PCC fails. Equivalent of the countable axiom of choice?

So there exist sequence of non empty sets such that each sequence meets only finitely many of Xn. Now we define sequence X = Xn*{n) with metric d[(x,m), (y,n)] = 0 if n=m and 1 otherwise.

(It works in pseudometric space, can we use same set with slight change in metric for metric space case)