proof for one equivalent statement of Baire category theorem

First, let’s assume Baire’s category theorem and prove the alternative statement.
We have $B={\bigcup }_{n=1}^{\mathrm{\infty }}{B}_n$, with $\text{int}(\overline{B_k})=\mathrm{\varnothing }\forall k\in 𝐍$.
Then

$$X=X-\text{int}(\overline{B_k})=\overline{X-\overline{B_k}}\forall k\in 𝐍$$

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

$$\bigcap _{n=1}^{\mathrm{\infty }}(X-\overline{B_n})=X-\bigcup _{n=1}^{\mathrm{\infty }}\overline{B_n}$$

is dense in $X$.But $B\subset {\bigcup }_{n=1}^{\mathrm{\infty }}\overline{B_n}⟹X-{\bigcup }_{n=1}^{\mathrm{\infty }}\overline{B_n}\subset X-B$, and then$X=\overline{X-{\bigcup }_{n=1}^{\mathrm{\infty }}\overline{B_n}}\subset \overline{X-B}=X-\text{int}(B)⟹\text{int}(B)=\mathrm{\varnothing }$.

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

Then $X-\overline{{\bigcap }_{n=1}^{\mathrm{\infty }}{B}_n}=\text{int}(X-{\bigcap }_{n=1}^{\mathrm{\infty }}{B}_n)=\text{int}({\bigcup }_{n=1}^{\mathrm{\infty }}X-B_n)=\mathrm{\varnothing }$ due to our hypothesis. Hence Baire’s category theorem holds.
QED