For sets A and B in a topological space X, the following hold:
(1) Cl(X-A)=X-Int(A)
(2) Int(A)∩ Int(B) =Int (A∩B)
If $\displaystyle x\in \text{int}(A) \cap \text{int}(B)$ then $\displaystyle x\in \text{int}(A) \text{ and }x\in \text{int}(B)$.
This means there are open subsets $\displaystyle U,V$ so $\displaystyle x\in U\subseteq A \text{ and }x\in V\subseteq B$.
Thus, $\displaystyle x\in U\cap V \subseteq A\cap B$, remember that $\displaystyle U\cap V$ is open subset.
Thus, $\displaystyle x\in \text{int}(A\cap B)$.
If $\displaystyle x\in \text{int}(A\cap B)$ then $\displaystyle x\in U\subseteq A\cap B$ for some open subset $\displaystyle U$.
But then $\displaystyle x\in U\subseteq A \text{ and }x\in U\subseteq B$.
Therefore, $\displaystyle x\in \text{int}(A) \text{ and }x\in \text{int}(B) \implies x\in \text{int}(A)\cap \text{int}(B)$.
Thus, $\displaystyle \text{int}(A\cap B) = \text{int}(A)\cap \text{int}(B)$.
Let $\displaystyle x\in \overline{X-A}$ then it means for any open set $\displaystyle U$ with $\displaystyle x\in U$ we have that $\displaystyle U\cap (X-A)\not = \emptyset$. But then there is not open set $\displaystyle U$ with $\displaystyle x\in U$ so that $\displaystyle U\subseteq A$ (otherwise the intersection would be empty). Therefore, $\displaystyle x$ is not and interior point of $\displaystyle A$, thus, $\displaystyle x\in X - \text{int}(A)$. Conversely, let $\displaystyle x\in X - \text{int}(A)$, this means $\displaystyle x\in X$ and $\displaystyle x\not \in \text{int}(A)$. But if $\displaystyle x\not \in \text{int}(A)$ it means for any open set $\displaystyle U$ so that $\displaystyle x\in U$ we have that $\displaystyle U\not \subseteq A$. Thus, for any open set $\displaystyle U$ with $\displaystyle x\in U$ we see that $\displaystyle U\cap X - A\not = \emptyset$ which means $\displaystyle x$ lies on the closure of $\displaystyle X-A$.