If x is in the closure of the closure of A, then it is in the closure of A I am trying to prove the following statement: if $x$ is in the closure of the closure of $A$, then it is in the closure of $A$ ($\bar{A}$).
Suppose $x$ is in the closure of the closure of $A$. Then it is the case that, $\forall \epsilon>0$, $B(x, \epsilon) \cap \bar{A} \neq \emptyset$. Suppose $y$ is in this intersection. Since $y$ is in $\bar{A}$, and $y$ is in the open set $B(x, \epsilon)$, $B(x, \epsilon)$ must contain at least one point from $A$ (by the definition of the closure). So the $B(x, \epsilon)\cap A\neq \emptyset$.
Is this correct? Thank you.
 A: Looks more or less good, yes! By more or less I mean if I was a TA and this was an answer on a test, I'd give it all points.

That said, I have some very minor and nitpicky comments strictly on how the proof is written.
Comment 1:
In particular, when you say

suppose $y$ is in this intersection.

It is clear from context what you mean by this intesection, but technically, you only wrote a statement true for all $\epsilon$. To make it more clear, I would add the sentence

"Let $\epsilon > 0$

before "suppose $y$ is in this intersection". So the beginning of the proof would be:

Suppose $x$ be an element of the closure of the closure of $A$. Further, let $\epsilon > 0$. Then, we know that $B(x,\epsilon)\cap \overline A\neq\emptyset$. Therefore, there exists some $y\in B(x,\epsilon)\cap \overline A$

Comment 2:
When you say "By the definition of closure", this is a little unclear because there are several equivalent definitions of closure. I assume you mean the one saying:

$B$ is the closure of $A$ if, for every $x\in B$ and every open set $O$ containing $x$, the intersection $O\cap A$ is nonempty.

If you mean that definition, then I would rewrite the end of the proof ever so slightly into:

Since $y$ is in $\bar{A}$, and $y$ is in the open set $B(x, \epsilon)$, we know that $B(x, \epsilon)\cap A$ is nonempty, which concludes the argument.

