The Cantor-Bendixson theorem states that
If $X$ is Polish then any closed subset of $X$ can be written as the disjoint union of a perfect subset and an at most countable subset.
It seems that we can weaken the condition as follows:
If $X$ is second countable (even not $T_1$), then any closed subset $F$ of $X$ can be written as the disjoint union of a perfect subset and an at most countable subset
The canonical proof applies as follows:
Suppose $\mathcal B$ is a countable base. Let $P$ be the set of condensation points, since $F$ is closed, $P\subseteq F$. For each $x\in F\setminus P$, we assign $B_x\in\mathcal B$ such that $x\in B_x$, and $B_x\cap F$ is at most countable, then there's a map $\phi\colon F\setminus P\to\mathcal B,x\mapsto B_x$. Note that the preimage of each $B_x$ is at most countable, therefore $F\setminus P$ is countable.
It remains to show that $P$ is perfect. Every point $p\in P$ is a condensation point of $F$, hence $P$, since $F\setminus P$ is countable, hence a limit point of $P$. Conversely, if $x\not\in P$, there's an open set $\mathcal O$ contains $x$ such that $\mathcal O\cap F$ is at most countable, hence $\mathcal O\cap P=\emptyset$.
Is it right?