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 $x$.