Does Cauchy Completeness imply the Heine-Borel theorem generally? I've been working through some reverse math with the completeness definitions of a metric space.  More over, I've learned that in a metric space X that is ordered, The Least Upper Bound Property, Cauchy Criterion, Nested Interval Theorem, and Heine-Borel Theorem are all equivalent (provided that the Archimedean Property is true).

My Question: Let X be a metric space and suppose X is Cauchy complete.  Does the Heine-Borel Theorem follow?  In other words, is it true that if every Cauchy sequence converges to a limit in X, then every closed and bounded set in X is compact?

I've been able to show this is true for $\mathbb{R}^n$.  Is this true for a general metric space?
 A: No.  Any infinite-dimensional Banach space is a counterexample to this.  The closed unit ball in such a space is closed and bounded but not compact. 
A: You already got a counterexample in another answer but let me add a potentially useful fact. If $X$ is a metric space then $Y\subseteq X$ is compact iff it is complete and totally bounded. In particular if $X$ is complete then $Y\subseteq X$ is compact iff it is closed and totally bounded. (Note that the closed unit ball in an infinite dimensional Banach space is bounded but not totally bounded)
A: A simple counter-example is the discrete metric, that is, $d(x,y)=1$ when $x\ne y,$ on an infinite set $X.$ Any $d$-Cauchy sequence is eventually constant and hence convergent. And  every subset of $X$ is closed and bounded, but only the finite subsets of  X are compact.
Another impediment is that if $d$ is a $complete$ metric on a set $X$ (that is, if every $d$-Cauchy sequence converges in $X$) then the metric $e(x,y)=\min (1,d(x,y))$ is also  complete, and $e$ generates the same topology as $d$ does.  Hence $d$ and $e$ generate the same collections of closed sets and of compact sets. So $X$ itself is closed and $e$-bounded, but not necessarily compact.
For example let $X=\Bbb R$ and $d(x,y)=|x-y|$ and $e(x,y)=\min (1,|x-y|).$
