How to prove that $\left(\overline{A}\right)'\subset A'$ in a Hausdorff space If $E$ is a Hausdorff space, and $A\subset E$ how to prove that $\left(\overline{A}\right)'\subset A'$ ?

We say that $x\in A'$ if and only if $\forall V\in \mathcal{V}_x,
 (V\setminus\{x\})\cap A\neq\emptyset$
We say that $x\in \overline{A}$ if and only if $\forall V\in
 \mathcal{V}_x, V\cap A\neq\emptyset$

So let $x\in \left(\overline{A}\right)' $ then $\forall V\in \mathcal{V}_x, (V\setminus\{x\})\cap \overline{A}\neq \emptyset$ 
then there exist $y$ such that $y\in V\setminus\{x\}$ and  $y\in \overline{A}$ so $y\in V$ and $\forall W\in \mathcal{V}_y, W\cap A\neq \emptyset$ 
We have that $E$ is a Hausdorff space so there exist a nbh $V_1$ to $x$ and  $V_2$ for $y$ such that $V_1\cap V_2= \emptyset$
But how we prove that  $(V\setminus\{x\})\cap A\neq \emptyset$
how to continue please
 A: Perhaps it is easier to prove the contrapositive: $(A')^c \subset ((\overline{A})')^c$.
Suppose that $x \in (A')^c$, so $x$ is not a limit point of $A$. Therefore there exists $V \in \mathcal V_x$ such that $W = V \setminus \{x\}$ does not intersect $A$. If we can show that $W$ does not intersect $\overline A$ then we're done.
We already know that $W$ does not intersect $A$, so it suffices to show that $W$ does not contain any limit points of $A$.
Note that $\{x\}$ is closed since the space is Hausdorff (see proof below), so $W = V \setminus \{x\} = V \cap (\{x\})^c$ is an intersection of two open sets, hence open. Suppose $y \in W$. If $y$ were a limit point of $A$, then $W$ would have to contain some $a \in A$, but this cannot happen because $W \cap A = \emptyset$.

Proof that all singletons are closed in a Hausdorff space: let $\{x\}$ be any singleton and $y$ be any other point. Then there is an open neighborhood $U_y$ which contains $y$ but does not contain $x$. Then $$U = \bigcup_{y \in E \setminus \{x\}} U_y$$ is a union of open sets, hence open, and $U$ contains every point of $E$ except $x$, hence $U = \{x\}^c$. Since $\{x\}^c$ is open, it follows that $\{x\}$ is closed.
A: Note that $\overline A = A \cup A'$. Since the derived set of a finite union is the union of the derived sets, we have $\left(\overline A\right)' = A' \cup A''$. Since $A'$ is closed in a $T_1$-space (see Set of limit points of a subset of a Hausdorff space is closed.), we have $A''\subseteq A'$ and thus $\left(\overline A\right)'\subseteq A'$.
A: Let $x\in\left(\overline{A}\right)'$ and let $V$ be an open set with $x\in V$.
Then $(V\setminus\{x\})\cap\overline{A}\neq\varnothing$. Let $y\in(V\setminus\{x\})\cap\overline{A}$.
In order to prove that $x\in A'$ it is enough to show that $(V\setminus\{x\})\cap A\neq\varnothing$.
We will do that by deducing a contradiction on base of the assumption: $(V\setminus\{x\})\cap A=\varnothing$. 
Note that $V$ is an open set that contains $y\in\overline{A}$, so that $V\cap A\neq\varnothing$.
$(V\setminus\{x\})\cap A=\varnothing$ combined with $V\cap A\neq\varnothing$ implies $V\cap A=\{x\}$. 
We have $y\in V\setminus\{x\}$ so that $y\neq x$. 
Let $U$ be an open set with $y\in U$ and $x\notin U$. Such set exists because the space is Hausdorff. 
(Actually it is enough allready if the space is T1.)
Then $y\in U\cap V$, $U\cap V$ is open and $(U\cap V)\cap A=\varnothing$. 
This contradicts that $y\in\overline{A}$ and we are ready.
