I have been asked (by a book) to prove the following:
In a $T_1$ space, the set of accumulation points of a subset is closed.
My concern is that the proof I have written seems to work in any topological space. Please help me find my mistake, but please don't provide a proof of the statement.
Let $X$ be a topological space. Let $A$ be a subset of $X$ and let $A'$ be the set of accumulation points of $A$. Let $x$ be an accumulation point of $A'$. If we let $N(x)$ be an open neighborhood of $x$ then we are guaranteed a point $a' \in N(x) \cap A'$. ($N(x)$ is open WLOG because every neighborhood contains an open neighborhood.) However, since $N(x)$ is open, this implies that $N(x)$ is also a neighborhood of $a'$. Since $a'$ is an accumulation point of $A$, this means that there is a point a of $A$ in $N(x)$, which implies that x was in fact an accumulation point of $A$, and therefore $x \in A'$. Since $A'$ contains all of its accumulation points, it must be closed.