I was trying to prove the following statement:
Let $(X,\mathscr{T})$ be a topological space. If the set of accumulation points of $\{x\}$ is closed for every $x\in X$, then the set of accumulation points of each subset of $X$ is closed.
I've tried to start with, for a subset $S\subset X$, $x\in (S')'\setminus S'$, where $S'=\{\mbox{accumulation points of }S\}$, and get a contradiction, but all i've showed is that $x\in S$ and $\{x\}'\subset S'$ (but doesn't seem to help...): You can choose an open set $V$ containing $x$ and such that $V\cap \left(S\setminus\{x\}\right)=\emptyset$ (because $x\notin S'$). Then $\exists y\in V\cap S'$, so $V$ is a neighborhood of $y$, and then $V\cap (S\setminus\{y\})\ne\emptyset$. This implies necessarily that $V\cap (S\setminus\{y\})=\{x\}$ and $x\in S$.
If $y\in\{x\}'$, then every open neighborhood $U$ of $y$ is a open neighborhood of $x$ and, repeating the argument, $U\cap S\setminus\{y\}$ is non-void, so $y\in S'$.
Please, any hint? Thanks!