Comparing definitions of closure I have come across the following definitions of closure

*

*$\overline{A}=A' \cup A$ is called closure, if $A'$ is the set of all limit points of A

*$\overline{A}=\cap _{\forall i \in \Lambda } V_i$, where $V_i$ is a closed set containing $A$
My understanding is that if these statements mean the same then $$A' \cup A= \cap _{\forall i \in \Lambda } V_i$$
under the same conditions in the definitions.
I tried to prove this as follows
Proof
If $A$ is closed then it contains all its limit points. In other words, $A'\subset A \Rightarrow \overline{A}=A'\cup A=A $. Similarly, it is clear that $ \cap _{\forall i \in \Lambda } V_i=A$. Thus, $$A' \cup A= A= \cap _{\forall i \in \Lambda } V_i$$
Suppose $A$ is not closed. Then $A$ does not contain all its limit points. If $a$ is a limit point of $A$, that is let $a \in A'$. Then it is also a limit point of $V_i \supset A$. This means that $a \in V\setminus A$ because $a \notin A$. Thus, $a \in \cap V_i\setminus A$. We can conclude that $A' \subset \cap (V_i\setminus A) $
We take not of the fact that $\cap _{\forall i \in \Lambda } V_i=A \cup \left( \cap (V_i \setminus A)\right)$. Thus, there is also need to show that $A' \supset \cap (V_i\setminus A) $. This is where I am failing to move on.
Can someone be of assistance?
 A: Let $V=\cap V_i$. As an intersection of closed sets, it must be closed.
By way of contradiction, let $b\in V$ such $b\notin A'\cup A$. Then there is some open neighbourhood $B$ of $b$ that does not intersect $A$. As $B$ contains no point of $A$, its complement $B^C$ must contain all points of $A$, so $B^C \supseteq A$. As $B$ is open $B^C$ must be closed*. Then as an intersection of closed sets $B^C\cap V$ is closed. It contains $A$, so it must be one of the $V_i$, say $V_j$. Then
$$V = \cap V_i = V_j\cap (\cap V_i) = V_j\cap V = (B^C\cap V)\cap V \subseteq B^C$$
but this is a contradiction as $b\in V$ but $b\notin B^C$. Thus $A'\subseteq \cap V_i$.
*You may have proven this already. If not, feel free to do so.
A: We know that $\overline A$ is closed because its complement $(A\cup A')^C=A^C\cap A'^C$ is the set of points that are neither in $A$ nor limit points of $A$. If this set is empty it is open and, if not, then for each point $p$ there is an open neighborhood that contains $p$ and does not intersect $A$. Therefore, the complement of $\overline A$ is open and $\overline A$ is closed.
Consider $\cap V_i$ where each $V_i$ is closed and contains $A$. Since each $V_i$ is closed it must also contain $A$'s limit points so $\overline A\subset V_i$ for all $i$. Therefore, $\overline A\subset\cap Vi$. But the intersection of all of the $V_i$ must also be a subset of each of the $V_i$ by definition and, since $\overline A$ is one of them, $\cap V_i\subset\overline A$ and therefore, $\overline A=\cap Vi$.
