We know that every ferfect set $E$ in a complete metric space $X$ is uncountable. My question is if there exists a metric space which is not complete, but every ferfect set in it is uncountable. The ferfect set here means a closed set which has no isolated points.
It is not so easy to construct an example of this kind, I think, because of the Hurewicz theorem: if $X$ is a coanalytic separable metrizable separable space, then either it is Polish (in particular, its perfect subspaces are uncountable) or it contains a closed subset homeomorphic to the rationals (so, a countable perfect closed subset). This is corollary 21.21 in Kechris's book "Classical Descriptive Set Theory".
Allowing for the axiom of choice, a Bernstein set will provide a counterexample. This is a subspace $A$ of the real line built by transfinite recursion in such a way that both $A$ and its complement intersect any nonempty perfect subspace of the reals. (this construction is also explained by Kechris, 8.24).
EDIT: my earlier argument was incomplete, as pointed out by Andreas Blass, so I'm following his idea below to show that a Bernstein set is indeed a counterexample.
If $C$ were a countable closed perfect subset of $A$, then the closure of $C$ would be perfect and closed in $\mathbb R$, hence uncountable. So $\overline C \setminus C$ is an uncountable Borel set contained in the complement of $A$; since an uncountable Borel set contains a Cantor set, the complement of $A$ must then contain a Cantor set, which is impossible since $A$ is a Bernstein set.
Thus $A$ does not contain any countable closed perfect set, and $A$ is certainly not completely metrizable since $A$ is not even Borel.
It seems plausible to me that, if you want an example which is a subset of $\mathbb R$, then you need some form of the axiom of choice, though I have not thought it through carefully; maybe the residing set theorists will know just how much choice is needed to answer this question...