Show that $(cl(A))'=A'$ where $A\subseteq E$, $(E, d)$ is a metric space and $d$ is the metric function. I am trying to prove that $(cl(A))'=A'$ in a metric space $(E, d)$ where $d$ is the metric function.
I'll explain my notation. If $A\subseteq E$, $x\in E$ is an accumulation point of $A$ if and only if for every open set $S\subseteq E$ (containing $x$), $S$ and $A$ have in common at least one different point of $x$, i.e., if and only if $S\setminus\{x\}\cap A\neq\emptyset$. The set of accumulation points of $A$ is denoted by $A'$ and is called derived set. On the other hand, the set $cl(A)$ is called the closure of $A$ and is defined by $cl(A)=A\cup A'$.
I have shown that $A'\subseteq (cl(A))'$ by virtue of $A\subset cl(A)\implies A'\subseteq (cl(A))'$ (using the fact $A\subseteq B\implies A'\subseteq B'$). Nevertheless, I'm stuck in the second inclusion. I have written the following: $x\in (cl(A))'\implies(S\setminus\{x\}\cap cl(A)\neq\emptyset\ \forall S$ environment of $x$ (open set containing $x$), i.e., $\left(S\setminus\{x\}\cap A\right) \cup \left(S\setminus\{x\}\cap A' \right)\neq\emptyset$. I would like to show that $\left(S\setminus\{x\}\cap A\right)$ is necessarily different from empty set but it has not been possible for me to do so. Any suggestions or help will be valuable. Thank you for your attention and have an excellent day.
Remark I can use the fact: $x\in A'\implies S\setminus\{x\}\cap A$ is an infinite set $\forall\ S$ environment of $x$.
 A: Note that for a metric space $(X,d)$, $x\in A'$ if and only if for every $\epsilon>0$, there is a $y\in A\setminus\{x\}$ for which $d(x,y)<\epsilon$.
If $x\in cl(A)'$ and $x\in A$, then for each $n$, there is a $y_n\in cl(A)\setminus\{x\}$ for which $d(y_n,x)<1/2n$. For each $y_n$, there is an $x_n\in A$ for which $d(y_n,x_n)<d(y_n,x)$ as $y_n\in A'$. Note that $x_n\neq x$ and possibly $x_n=y_n$. Then by the triangle inequality we have 
$$d(x_n,x)\leq d(x_n,y_n)+d(y_n,x)<1/n.$$ 
Hence, $x\in A'$.
If $x\in cl(A)'$ and $x\notin A$, then there is some open set $S$ containing $x$ which intersects $cl(A)\setminus\{x\}$ and hence either $A\setminus\{x\}=A$ or $A'\setminus\{x\}$. If $S$ does not intersect $A$ then there is some $y\in A'\setminus\{x\}\cap S$. But then $S$ is an open set containing $y$ which does not intersect $A$, which is a contradiction as $y\in A'$. Thus, $S$ must intersect $A$ so $x\in A'$.
A: If $p$ is any member of $A'$ and if $U$ is any open set containing $p$ then there exists $q\in U\cap A$ with $q\ne p.$ Since $A\subset \bar A$ we have $p\ne q\in U\cap \bar A.$ So $p\in (\bar A)'.$
Therefore $A'\subset (\bar A)'.$
If $p$ is any member of $ (\bar A)'$ and if $U$ is any open set containing $p$ then there exists $q\in  U\cap \bar A$ with $q\ne p.$  There exists an open set $V$ such that $q\in V$ and $p\not \in V .$ Now $U\cap V$ is an open set containing $q,$ and  $q\in \bar A,$ so there exists $r\in A\cap (U\cap V).$ We have $r\in U\cap A$ and $r\ne p$ ( because $r\in V$ and $p\not \in V$). So $p\in A'.$
Therefore $(\bar A)'\subset A'.$
This is valid in any $T_1$ space $E$.
