$x\in \bar{A}$ iff every neighbourhood of $x$ intersects $A$. 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 $\bar A$.

 A: $x \not \in \overline A$ 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 $A \iff x \in \overline A$.
A: You define $\bar{A}$ as the least closed set containing $A$ (because the intersection of closed sets is closed).
Thus a point $x$ is not in $\bar{A}$ if and only if there is a closed set $C$ such that $A\subseteq C$ and $x\notin C$.
Now note that $x$ not belonging to a closed set $C$ containing $A$ is the same as $x$ belonging to an open set $U$ (precisely $U=X\setminus C$) whose complement contains $A$, that is, not intersecting $A$. An open set containing $x$ is a neighborhood of $x$.
This proves one direction. For the converse, if $V$ is a neighborhood of $x$ not intersecting $A$, $V$ contains an open set $U$ with $x\in U$, by definition of neighborhood. Since $V\cap A=\emptyset$, also $U\cap A=\emptyset$. Set $C=X\setminus U$: then $C$ is a closed set containing $A$ and $x\notin C$.
A: By definition, $\bar{A}$ is the smallest closed set containing $A$.  Let $x\in\bar{A}$ and $V$ an open set containing $x$.  Then $V^c$ is closed so $\bar{A} \cap V^c$ is closed.  Since $\bar{A} \cap V^c \neq \bar{A}$ (why?), we have that $\bar{A} \cap V^c$ is properly contained in $\bar{A}$.  
By minimality, $A$ is not contained in $\bar{A} \cap V^c$, and in particular there's some $a \in A$ so that $a \notin V^c$, so $A \cap V \neq \emptyset$.
A: Contradiction both ways.
($\Rightarrow$) The closure of $A$ is a closed set that is a subset of every closed set which contains $A$. If $x$ is in the closure of $A$, then $x$ is in every closed set containing $A$. If $x$ is in an open set that does not intersect $A$ then take the complement of this open set. That complement is closed and contains $A$ hence giving a contradiction since $x$ was assumed to be in the closure of $A$. 
($\Leftarrow$) Assume every neighborhood of $x$ intersects $A$. Take a closed set containing $A$. If $x$ was not in this closed set, then $x$ is in its complement which is an open set that does not intersect $A$ which is a contradiction.
A: Answer: I think this can be best "visualized" if you consider X as a topological space where the topology is induced by a metric, e.g. $X = \mathbb{R}^2$ with the topology induced by the standard Euclidean metric (i.e $\mathbb{R}^2$ with the topology that is formed by open $\epsilon$-balls). Then, the closure $\bar A$ of a set $A$ contains "all points that are arbitrarily close to points of $A$", including points in $A$ itself. In other words, $\bar A$ contains $A$ and all its boundary points. If you take for example $A$ as the open unit ball $B_1(0) \subset \mathbb{R}^2$ centered at $0$, then the "points that are arbitrarily close to $B_1(0)$" are exactly the ones that lie on the unit circle centered at $0$.
What Munkres now proves is the contraposition of what I said above: He shows that a point $x \in X$ that is not arbitrarily close to $A$ can't be contained in $\bar A$, and vice-versa. If your $x$ is not arbitrarily close to $A$, then you can always find a mini-mini open $\epsilon$-ball $B_\epsilon(x)$ centered at $x$ such that no points of $B_\epsilon(x)$ lie inside $A$, i.e. $B_\epsilon(x)$ doesn't intersect $A$. On the other side, if you have a point $x$ and an open $\epsilon$-ball $B_\epsilon(x)$ that doesn't intersect $A$, then $x$ can't be arbitrarily close to $A$, hence $x \notin \bar A$.
You can also state the theorem intuitively without using contraposition: If you have a point $x$ such that every open $\epsilon$-ball $B_\epsilon(x)$ around $x$ intersects $A$ no matter how small $\epsilon$ is, then $x$ must be arbitrarily close to $A$, hence $x \in \bar A$. And if $x \in \bar A$, then $x$ is arbitrarily close to $A$, hence every open $\epsilon$-ball $B_\epsilon(x)$ must intersect $A$. 
This description translates to arbitrary topological spaces where the open $\epsilon$-balls are just open sets defined by the topology on the space. However, you don't have always a metric on a topological space that gives you an exact notion of distance, but in fact the topology gives an abstract sense of "distance" by defining open sets.
I hope that my words could "catch" a bit the intuition about the theorem.
Note: The reason why Munkres proves the theorem by contraposition is that he only has to find one open neighborhood of a point $x \notin \bar A$ that doesn't intersect $A$ to conclude the $\Rightarrow$-direction, whereas without contraposition he has to show that every possible open neighborhood of $x \in \bar A$ has to intersect $A$ to conclude the $\Rightarrow$-direction, and there can be quite a lot of open neighborhoods around a point $x \in \bar A$, depending on the topology.   
