If $E$ is a perfect subset of $X$ and $A$ is an open subset of $X$, then $\overline{E ∩ A}$ is perfect. 
Let $X$ be a metric space. Show that if $E$ is a perfect subset of $X$ and $A$ is an open subset of $X$, then $\overline{E ∩ A}$ is perfect.

Here $\overline{E ∩ A}$ is closure of $E ∩ A$. Also, $Y$ is perfect $Y$ is closed and if every point of $Y$ is a limit point of $Y$.
My try of proof: 
Suppose $\overline{E ∩ A}$ is not perfect. Then since $\overline{E ∩ A}$ is closed, there exists a $p\in \overline{E ∩ A}$ such that $p$ is not limit point of $\overline {E ∩ A}$. We see that $p\in (E ∩ A)'$ is impossible (where $(E ∩ A)'$ is set of all the limit points of $E ∩ A$), because, by exercise 6, ch 2, in Rudin's Principles of Mathematical Analysis, we know that $E ∩ A$ and $\overline{E ∩ A}$ have the same limit points, so if $p\in (E ∩ A)'$, then $p$ was limit point of $\overline{E ∩ A}$ as well.  I have problem here: If $p\in E ∩ A$, since every point in $E$ is limit point, because $E$ is perfect, ...... I want to show that $p$ is limit point of $\overline{E ∩ A}$  as well.    Also, I can't see where in proof the fact that $A$ is open is needed.
 A: For convenience I'll use the following fact: In a metric or even already in a $T_1$ space, $x$ is a limit point of $B$ iff every open neighbourhood of $x$ contains infinitely many points of $B$. The rest is simply definition chasing:
Let $p$ be a point of $\overline{E \cap A}$. 
We need to show it's a limit point of $\overline{E \cap A}$. 
So let $O$ be any open set that contains $p$. 
By the definition of closure there is some point $q \in O \cap E \cap A$. As $A$ is open so is $O \cap A$. (Here I use $A$ is open)
As $O \cap A$ is an open set containing $q \in E$, which is perfect, $O \cap A$ contains infinitely many points of $E$.  It's clear these points in $O \cap A \cap E$ are also in $\overline{E \cap A}$. So $O$ contains infinitely many points of $\overline{E \cap A}$ as required.
So $\overline{E \cap A}$ is closed and every point of it is a limit point of it. So it's perfect.
A: The problem can be phrased in the following general setting:

let $(X, \mathscr{T})$ be a $\mathrm{T_1}$-space, and $M \subseteq X$ a
dense-in-itself subset. If $U \in \mathscr{T}$ is an open set, then $\overline{M \cap U}$ is perfect.

Recall that a subset $M$ is said to be dense-in-itself if it has no isolated points. This is expressed as $M \subseteq M'$ and is equivalent to $\overline{M}=M'$ without any hypothesis whatsoever on the topological space at hand.
A neat argument can be made in the sequence of the following propositions:


*

*In an arbitrary topological space $(X, \mathscr{T})$, if $M\subseteq X$ is dense-in-itself and $U$ is an open subset (a member of $\mathscr{T}$) then $U\cap M$ is also dense-in-itself.


Proof : let $x \in U\cap M$ be arbitrary and let $V \in \mathscr{T}$ be an arbitrary (open) neighbourhood of $x$. Then $U\cap V$ is an open set containing $x$, in other words an open neighbourhood of point $x$. As $M$ is dense-in-itself, $x$ is also an accumulation point of $M$ and thus we must have by definition that $(V\cap U)\cap M\setminus\{x\} \neq \emptyset$. Since set intersection is associative, the previous relation can just as well be expressed as $V\cap ((U\cap M)\setminus \{x\})\neq \emptyset$, which tells us that $x$ is adherent to $U\cap M\setminus \{x\}$ and thus an accumulation point of $U\cap M$. The relation $U\cap M \subseteq (U\cap M)'$ thus follows. $\Box$



*In a $\mathrm{T_1}$-space $(X, \mathscr{T})$, if $M \subseteq X$ is dense-in-itself then its closure is perfect.


Proof : recall that in general (without any assumptions on the space considered) a subset $M$ is called perfect if $M=M'$; this is equivalent to claiming that $M$ is closed and dense-in-itself. Also, let us recall that in any $\mathrm{T_1}$-space one has $S''\subseteq S'$ and hence $\overline{S'}=S'=(\overline{S})'$ for any subset $S$. Applying these equalities to $M$ in particular we obtain $M'=(\overline{M})'$ on the one hand and $M'=\overline{M}$ on the other, as $M$ is dense-in-itself. Hence, $\overline{M}$ equals its own derivative and is thus perfect. $\Box$
A: The other answers have proved the result under the assumption of a $T_1$ space.  The result is actually true in any topological space without separation assumptions.
Recall these definitions:

*

*A subset $A$ of a space $X$ is called dense-in-itself if $A$ has no isolated
point in $A$.  Equivalently, all the points of $A$ are limit points of $A$.

*A subset $A$ of a space $X$ is called a perfect set if $A$ is closed and
dense-in-itself.  In other words, $A$ is closed without isolated point.


Proposition: Let $X$ be a topological space.  If $O\subseteq X$ is an open set and $A\subseteq X$ is dense-in-itself, then $O\cap A$ is dense-in-itself.

Proof: Suppose by contradition that $O\cap A$ has an isolated point $x$.  Then there is an open set $U\subseteq X$ such that $U\cap O\cap A=\{x\}$.  But since $U\cap O$ is an open nbhd of $x$, this shows that $x$ is an isolated point of $A$.  So $A$ is not dense-in-itself.

Proposition: Let $X$ be a topological space.  If $A\subseteq X$ is dense-in-itself, then $\overline A$ is a perfect set.

Proof: $\overline A$ is closed, so we only need to show that $\overline A$ has no isolated point.  Suppose $x$ is an isolated point of $\overline A$.  Take an open set $U\subseteq X$ with $U\cap\overline A=\{x\}$.  Since $U$ is a nbhd of $x$ and $x\in\overline A$, the set $U\cap A$ is nonempty, and is also a subset of $U\cap\overline A$.  Hence $U\cap A=\{x\}$, showing that $x$ is an isolated point of $A$ and contradicting the fact that $A$ is dense-in-itself.
Putting the two propositions above together shows the slightly more general result:

Corollary: Let $X$ be a topological space.  If $A\subseteq X$ is dense-in-itself and $O\subseteq X$ is open, then $\overline{A\cap O}$ is a perfect set.

