How to prove this property on interior and closure of sets? $\newcommand{\inr}{\operatorname{int}}$

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

Proof : 
$(\implies)$ I suppose that there exists a set $B\in P(X)$ with $\overline{B}=X$ and $B \cap A = \emptyset$, then $B \subseteq A^{\complement}$, using some properties of closure sets i get:
$X=\overline{B}\subseteq\overline{A^{\complement}}=(\inr(A))^{\complement}$
then $X=\inr((A))^{\complement}$, so $\inr(A)$ needs be the empty set.
But, I don't know how to show the $(\impliedby)$ part.
 A: In any space $X$ and any $B\subset X$ the 3 sets $int(B),\,int(B^c), $ and $\partial(B)=\partial (B^c)=\overline B\cap \overline {B^c}, $ are pairwise disjoint and their union is $X.$
The $(\leftarrow)$ part is equivalent  to: If $int(A)=\emptyset$ then there exists a dense $B\subset X$ such that $A\cap B=\emptyset$.
If $int(A)=\emptyset$ let $B=A^c.$ Then (obviously) $A$ is disjoint from $B.$ And $B$ is dense in $X$ because $$\overline B\supset int(B)\cup \partial B=$$ $$=int(B)\cup \partial B\cup\emptyset=$$ $$=int(B)\cup\partial B\cup int (A)=$$ $$=int(B)\cup\partial (B)\cup int(B^c)=X.$$
A: A contrapositive proof.
Assume int A empty.
Then X = $\overline {X - A}$
and (X - A) $\cap$ A is empty.
A: You seemingly know that $$\overline{A^\complement}=\operatorname{int}(A)^\complement\tag{1}$$
and this implies that $$\operatorname{int}(A) = \emptyset \iff \overline{A^\complement}=X\tag{2}$$
And that last equation shows the way:
If $A$ has the property that it intersects all dense sets $B$, it has non-empty interior because otherwise its complement $A^\complement$ is dense and disjoint from it, contradicting the property.
For the forward direction it suffices to note that if $A$ has non-empty interior, it contains a non-empty open set and all non-empty open sets intersect all dense sets $B$ and so $A$ does too, a fortiori.
