Definition: The closure of a set $A$ is $\bar A=A\cup A'$, where $A'$
is the set of all limit points of $A$.
Claim: $\bar A$ is a closed set.
Proof: (my attempt) If $\bar A$ is a closed set then that implies that it contains all its limit points. So suppose to the contrary that $\bar A$ is not a closed set. Then $\exists$ a limit point $p$ of $\bar A$ such that $p\not \in \bar A.$ Clearly, $p$ is not a limit point of $A$ because if it were then $p\in \bar A$. This means that $\exists$ a neighborhood $N_r(p)$ which does not contain any point of $A$. But $p$ is a limit point of $\bar A$ so it must contain an element $y\in \bar A-A$ in its neighborhood $N_r(p).$ Of course, $y$ is a limit point. Now, $0<d(p,y)=h<r.$ If we choose $0<\epsilon<r-h$, then $N_{\epsilon}(y)$ will contain no point $x\in A$, which is a contradiction since $y$ is a limit point.
I want to know if this proof is correct.
Best Answer
Your proof is correct, maybe we can make it slightly faster.
Let $z$ be a limit point of $\overline A$. Every open set $U$ containing $z$ must contain a point $x$ in $\overline A$. If this point $x$ is in $A^\circ$ then $U$ must intersect $A$ because it contains a limit point of $A$. If $x$ is in $A$ then $U$ obviously intersects $A$.
So every limit point of $\overline A$ is a limit point of $A$, and $\overline A$ contains all of its limit points.