First, let's assume Baire's category theorem and prove the alternative statement.
We have $B=\bigcup_{n=1}^\infty B_n$ , with ${int}(\overline{B_k}) = \emptyset \; \forall k \in \mathbf{N}$ .
Then $$ X=X-\textrm{int}(\overline{B_k})=\overline{X-\overline{B_k}}\;\forall k \in \mathbf{N $$ Then $X-\overline{B_k}$ is dense in $X$ for every $k$ . Besides, $X-\overline{B_k}$ is open because $X$ is open and $\overline{B_k}$ closed. So, by Baire's Category Theorem, we have that $$ \bigcap_{n=1}^\infty (X-\overline{B_n})=X-\bigcup_{n=1}^\infty \overline{B_n} $$ is dense in $X$ . But $B\subset\bigcup_{n=1}^\infty \overline{B_n} \Longrightarrow X-\bigcup_{n=1}^\infty \overline{B_n} \subset X-B$ , and then $X=\overline{X-\bigcup_{n=1}^\infty \overline{B_n}} \subset \overline{X-B}=X-{int}(B) \Longrightarrow {int}(B)=\emptyset$ .

Now, let's assume our alternative statement as the hypothesis, and let $(B_k)_{k\in N}$ be a collection of open dense sets in a complete metric space $X$ . Then ${int}(\overline{X-B_k})={int}(X-{int}(B_k))={int}(X-B_k)=X-\overline{B_k}=\emptyset$ and so $X-B_k$ is nowhere dense for every $k$ .

Then $X-\overline{\bigcap_{n=1}^\infty B_n} = {int}(X-\bigcap_{n=1}^\infty B_n) = {int}(\bigcup_{n=1}^\infty X-B_n) = \emptyset$ due to our hypothesis. Hence Baire's category theorem holds.
QED