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.