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?

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 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 $$e(x,y)=\min (1,|x-y|).$$