Alternative proof that the union of $A$ and $A'$ is closed for any set $A$ I want to see, if this alternative proof that $A \cup A'$ is closed for any set $A$ is correct. Standard references I checked after doing this problem contained a completely different proof, so I am not sure if what I have done is correct.
$A'$ here denotes the set of limit points of $A$.
Prove: $A \cup A'$ is closed for any set $A$.
Proof: Let $p$ be a limit point of $A \cup A'$. The result is proved if we show that $p$ is an element of $A \cup A'$. Since $p$ is a limit point of $A \cup A'$, $p$ is a limit point of $A$ or $p$ is a limit point of $A'$. If $p$ is a limit point of $A'$, then $p$ is an element of $A'$ (since $A'$ is closed) and so $p$ is an element of $A \cup A'$. If $p$ is a limit point of $A$, then by definition, $p$ is an element of $A'$ and so $p$ is an element of $A \cup A'$. Therefore $A \cup A'$ contains each of its accumulation points and hence is closed.

The proof is correct, but somewhat overly complicated. It's easy to see that $A\subseteq A'$ (consider constant sequences), so $A\cup A'=A'$ and you use in your proof that that is closed.
