Intersection of a Perfect and an open subsets of X In Prof. George Bergman's Real analysis supplementary exercises, question 2.2:10. It is required to proof that the closure of the intersection of a perfect set $E$ with an open set $A$ is again perfect, in some metric space $X$. 
I tried one method to attack the problem, to assume that for a point $p$ in ($E \cap A$) is not a limit point(so an isolated point), so there is a neighborhood of $p$, s.t it's intersection with the set ($E \cap A$) contains no points other than $p$, and this would lead to a contradiction, as the intersection is not empty, after going through some steps. Thus for all $p$ that belong to $E \cap A$ are limit points, and so with the addition of the set of limit points. It is perfect.
What if we imagine an open set in ${R}^2$ that resembles an open ball with a "hole" that contains a single point $m$. (both an interior and an isolated point of $A$) If $E \cap A$ contains $m$, and, so it cannot be a limit point in ($E \cap A$) so the closure of $E \cap A$ seizes to be perfect? 
I guess there is a gap in my understanding, so it'd be helpful if some hints/discussions are given to help fill in the gap- thank you.
 A: 
both an interior and an isolated point of $A$

An open set in $\mathbb R^2$ cannot have isolated points. An isolated point $m$ in $A$ forms a one-point 
open subset in the induced topology on  $A$. Since $A$ itself is open, $\{m\}$ would have to be open in $\mathbb R^2$,
which it is not.  
The argument in the preceding paragraph  shows that if an open subset $A$ of a metric space $X$ has an isolated point
$m$, then $\{m\}$ is open in $X$. Therefore, $m$ is an isolated point in $X$. It follows that any subset of $X$ containing $m$ will have $m$ as an isolated point. Put another way, a perfect subset of $X$ cannot contain $m$. 

Your approach to the problem is sound. Perhaps the following will help to complete it: since $p\in A$ and $A$ is open,
there is a neighborhood of $p$ (call it $U$) such that $U\subseteq A$. Also, if $p$ is an isolated point of $E\cap A$,
then there is a neighborhood of $p$ (call it $V$) such that $(E\cap A)\cap V=\{p\}$. Now look at $U\cap V$ and at 
its intersection with $E$. 
