What is the intuition behind the proof of this "one-step" topology theorem? I am not sure if I understand the definitions of interior and closure of a set. So I am asking for help in the form of an example:
$C(\cdot)$ means the complement of, $int(\cdot)$ means the interior of, $cl(\cdot)$ means the closure of. 
Theorem: $$C(int(A))=cl(C(A))$$
 A: Take the very simple example where $A$ is the unit disc (without the boundary) union the point $(0, 1)$.
$int(A)$ = the unit disc. $C(A) = \{ (x, y) | x^2+y^2 \geq 1\} - \{(0, 1)\}$.
$$C(int(A)) = cl(C(A)) = plane - unit\ disc.$$
In general, you can think of $cl(.)$ as including all points near the set and $int(.)$ as excluding all the points near the complement of the set.
A: The interior of $A$ is the union of all open subsets included in $A$.
So the complement of the interior of $A$ is the intersection of all the complement of open subsets included in $A$.
But the complement of the open subsets included in $A$ are exactly the closed subsets containing the complement of $A$.
Since the closure of the complement of $A$ is the intersection of all closed subsets containing the complement of $A$, it is equal to the complement of the interior of $A$.
A: Let us consider the notion of $\textit{interior point}$  for $B$  as a point having an open neighborhood included in B , and the notion of $\textit{accumulation point}$  for $B$  as a point without open neighborhoods disjoint from $B$.
Now what you write is: 
$p$ is not an interior point of $A$  if and only if $p$  is an accumulation of $C(A)$
