The closure of the interior of the boundary is a subset of the closure of the intersection between a set and the interior of the boundary The statement is the following:  

Let X be a topological space and let A be a subset of X. Then $ cl(int(\partial A)) \subseteq cl(A \cap int (\partial A) ) $

Notation: $cl$ means closure, $\partial $ means boundary of a set, and $int $ means interior of a set.
These are the posts I have read so far:
A set which the interior of its boundary is not empty
When is the closure of an intersection equal to the intersection of closures?
If the interior of the boundary of a set is nonempty, then the interior of that set is empty 
However I have not been able to prove it yet.
How can I prove that statement?
 A: To clarify:


*

*A neighbourhood of $a$ is an open set containing $a$.

*The interior of a set $A$ is the set of points which have a neighbourhood $B$ which is a subset of $A$.

*The closure of $A$ is the set of points for which all neighbourhoods intersect $A$.


The boundary is the set of points whose neighbourhoods always intersect $A$, but do not lie in the interior of $A$. 
Its interior is the set of points who have a neighbourhood which lies entirely in the boundary. As boundary points lie in the closure, this neighbourhood must also intersect $A$. 
The closure of the interior of the boundary are the points for which every neighbourhood intersects the interior of the boundary. 
Now assume $a \in cl(int(\partial A ))$. Let $U_a$ be one of its neighbourhoods; it intersects $int(\partial A)$ in at least one point $y$. Let $V_y$ be a neighbourhood of $y$ which lies completely in the interior boundary. The intersection $W_y := U_a \cap V_y$ is also a neighbourhood of $y$ which lies completely in the interior boundary; furthermore, it is a subset of $U_a$. As argued before, $W_y$ must contain a point $z$ of $A$. This point $z$ is also a point of the interior boundary. Therefore, $z \in A \cap int(\partial A)$. 
We have just proven that every neighbourhood of $a$ intersects a point $z \in  A \cap int(\partial A)$. By definition it then follows that 
$$cl(int(\partial A )) \subset cl( A \cap int(\partial A) )$$
which concludes the proof.
