I read the proof of this theorem from Munkres, however I don't really understand intuitively why this is true. If someone could provide me of intuition of this theorem that would be nice. $\bar{A}$ is defined as the intersection of all closed sets that contain A.
Proof. $\ \ $ Consider the statement in $\rm(a)$. It is a statement of the form $P\Leftrightarrow Q$. Let us transform each implication to its contrapositive, thereby obtaining the logically equivalent statement $({\rm not}\, P)\Leftrightarrow({\rm not}\, Q)$. Written out, it is the following. $$x\notin \bar A\iff\text{there exists an open set $U$ containing $x$ that does not intersect $A$.}$$ $\quad$ In this form, our theorem is easy to prove. If $x$ is not in $\bar A$, the set $U=X-\bar A$ is an open set containing $x$ that does not intersect $A$, as desired. Conversely, if there exists an open set $U$ containing $x$ which does not intersect $A$, then $X-U$ is a closed set containing $A$. By definition of the closure $\bar A$, the set $X-U$ must contain $\bar A$, therefore, $x$ cannot be in $\bar A$.
Best Answer
$x \not \in \overline A$ means $x$ is not in the intersection of all closed sets containing $A$.
...means that there is a closed set $V$ such that $A \subset V$ but $x \not \in V$.
...means $x \in V^c= U$ which is open and $U \cap A = \emptyset$ (Because $A \subset V = U^c$). Thus there are open neighborhoods of $x$ that are subset of $U$ and therefore do not intersect $A$.
So if $x \not \in \overline A \implies $ there is an open neighborhood of $x$ that do not intersect $A$.
Conversely if the is an open neighborhood $N$ containing $x$ that doesn't intersect $A$ then $N^c$ is closed and $A \subset N^c$ and $a \not \in N^c$. So $a \not \in \cap \text{Closed Sets Containing A} = \overline A$.
so So if there is an open neighborhood of $x$ that do not intersect $A \implies$ $x \not \in \overline A$ .
So there is an open neighborhood of $x$ that do not intersect $A \iff$ $x \not \in \overline A$ .
Therefore: All open neighborhoods of $x$ intersect $A \iff x \in \overline A$.