To prove the statement, we need the two theorems:
T1. Let $X$ be a complete metric space and $A\subset X$. Then $A$, with the metric of $X$, is a complete metric space if and only if $A$ is closed in $X$.
T2. If $X$ is a (nonvoid) complete metric space having no isolated points and $E$ is a countable dense subset of $X$, then $E$ is not a $G_\delta$ set.
The statement:
If A is a perfect nonvoid subset of a complete metric space, then $A$ is uncountable. [Hint: Use T1 and T2].
I think I'm stuck because I cannot see the usefulness of $G_\delta$ sets.
If $A$ is perfect, then it is equal to its limit points (has no isolated points) and thus $A$ is closed. By T1, $A$ is complete. In particular, if $A$ is a (dense) countable subset, then by T2 $A$ is not a $G_\delta$ set.
This means that $A$ cannot be expressed as the countable intersection of open subsets, fine, but this should rise a contradiction that I cannot see.
Best Answer
$A$ is closed and therefore complete by the first theorem. Now work in the space $A$. If $A$ were countable, it would be a countable, dense, open subset of itself.