Let $A$ be a set and $A'=\{x: x \text{ is the accumulation point of A}\}$, which is called the derived set of $A$ and $\overline{A}$ closure of a set
I'm trying to prove that for a subset $A$ of a topological space $X$
$\overline{A'}=A'$
First $A'\subseteq \overline{A'}$
Using $B\subseteq \overline{B}$ for any subset in $X$, I prove it
But the $\subseteq$ part
How I can prove $\overline{A'}\subseteq A'$?
Any sugestion
Here a proof under the extra condition that singletons are closed. Actually I am not sure whether this condition can be missed.
Note that $A'\subseteq\overline{A}$ so that $\overline{A'}\subseteq\overline{A}$.
Based on $x\notin A'$ we will prove that $x\notin\overline{A'}$.
Let $x\notin A'$. Then $x\in U$ and $U\cap A=\emptyset\vee U\cap A=\left\{ x\right\} $ for some open set $U$. If $U\cap A=\emptyset$ then $x\notin\overline{A}$ and consequently $x\notin\overline{A'}\subseteq\overline{A}$, and we are ready.
Now the other possibility: $U\cap A=\left\{ x\right\} $. Assume that $x\in\overline{A'}$ . Then $U\cap A'\neq\emptyset$ so there must be some $y\in U\cap A'$. Here $x\notin A'$ tells us that $x\neq y$. Then $V:=U-\left\{ x\right\} $ is an open set (here it is used that singletons are closed) with $y\in V$ and $V\cap A=\emptyset$ contradicting that $y\in A'$.
This contradiction allows us to conclude that $x\notin\overline{A'}$.
