Prove every derived set is closed if this is the case for singleton sets Suppose that for each $x \in X$, the set of accumulation points of $\{ x \}$ is closed. Then for each $S \subseteq X$, the set of its accumulation points is closed.
This is the last part of exercise $D$ of chapter 1. I managed to solve all the previous points (not indicated here), but unfortunately this one remains open for me. Thank you in advance!

Suppose $S \subseteq X$ and $x \notin S'$. We need to find some open $U$ containing $x$ and such that $U$ misses $S'$. 
$x \notin S'$ means that there exists some open $O$ containing $x$ such that $O \cap S \subseteq \{x\}$.
If $O \cap S = \emptyset$, we can pick $U = O$.
If $O \cap S = \{x\}$, then use that $X \setminus \{x\}'$ is open...
