# How to prove this property on interior and closure of sets?

Assume that $$(X, \tau)$$ is a topological space, and $$A \in P (X)$$. Show that $$int(A) \neq \emptyset$$ if only if for all $$B \in P(X)$$ with $$\overline{B}=X$$ then $$B \cap A \neq \emptyset$$

For prove $$(\rightarrow)$$:

I suppose that exist a set $$B\in P(X)$$ with $$\overline{B}=X$$ and $$B \cap A = \emptyset$$, then $$B \subseteq A^{c}$$, using some properties of closure sets i get:

$$X=\overline{B}\subseteq\overline{A^{c}}=(int(A))^c$$

then $$X=int((A))^c$$, so $$int(A)$$ needs be the empty set.

But, i don’t know how to show the $$(\leftarrow)$$ part.