This question was a true /false question in my real analysis quiz today and I am unable to provide a reason for it.
Question is : A closed and bounded subset of complete metric space is compact.
I think it is not compact but only because any open cover may not have finite subcover but I am unable to provide a counterexample for the same.
For a counterexample, you can consider $[0,1]$ with the discrete metric $d(x,y) = 1$ if $x \ne y$, $d(x,y) = 0$ if $x = y$. This is complete because any Cauchy sequence is eventually constant and hence convergent, but the open cover $\{B(x,\frac 12) : x \in [0,1]\}$ has no finite subcovers because each ball $B(x, \frac 12) = \{x\}$ is a singleton.