Prove $W$ is closed iff for all $A\subset X$, $int(K\cup int(A))=int(K\cup A)$ Let $(X,\tau)$ be a topological space and $W\subset X$. Prove that if $W$ is closed, then for all $A\subset X$, $(W\cup A^\circ)^\circ=(W\cup A)^\circ$.
Here's what I have:
$A^\circ \subset A$, so $A^\circ\cup W\subset A\cup W$. Then $(A^\circ \cup W)^\circ \subset (A\cup W)^\circ$.
Now, let $p\in (A\cup W)^\circ$. Then there is $U\in \tau$ such that $p\in U\subset A\cup W$.
I want to find $V\in \tau$ such that $p\in V\subset W\cap A^\circ$.
First I noticed that $U=(U\cap W)\cup (U\cap (X\setminus W))$ and $(U\cap (X\setminus W))\in \tau$. 
I still can't manage to find the adequate $V$ to conclude that $p\in(W\cap A^\circ)^\circ$.
Any help would be appreciated.

It suffices to take $V:=U$. As you commented $U=(U-W)\cup(W\cap U)$ (Note that I am writing $U-Q=U\cap (X-W)$). Since $U\subset A\cup W$, then necessarily $U-W\subset A$. Besides, since $W$ is closed, then $U-W$ is open. Therefore $U-W\subset A^o$. 
Consequently, since $U-W\subset A^o$ and $W\cap U\subset W$, we finally obtain  $U=(U-W)\cup(W\cap U)\subset A^o\cup W$.
