Equivalent formulations of Baire's Lemma / Theorem I use the following notation $int_Z(A)$ for the interior of $A \subset Z$ in a metric space $Z$.
Now I have the following problem understanding a proof:

Let $(M,d)$ be complete. Then we have that statement i) implies ii).
i) If $M= \bigcup_{j=0}^{\infty}A_j$ with $A_j$ all closed, then there exists a $j_0$ with $int_M(A_{j_0}) \neq \emptyset$.
ii) If $int_M(\bigcup_{j=0}^{\infty}A_j) \neq \emptyset$ with closed $A_j$'s, then there exist a $j_0$ with $int_M(A_{j_0}) \neq \emptyset$

I came as close to understanding the proof as this:
Define the metric space $(X,d)=(\overline{B_{r}(x)},d)$ where $B_{2r}(x) \subset int(\bigcup_{j=0}^{\infty}A_j) \subset \bigcup_{j=0}^{\infty}A_j$ (since it has non-empty interior). Then by using (i) on the sets $X\cap A_j$ we get a $j_0$ with $int_{X}(X\cap A_{j_0})\neq \emptyset$.
Now how do I show that I have $int_M(A_{j_0}) \neq \emptyset$, if I only know that $int_{X}(X\cap A_{j_0})\neq \emptyset$?
Thanks a lot in advance!

Let $p\in int_X(X\cap A_{j_0})=S\cap X$ where $S$ is open in $M$.  Let $b>0$ such that $B_b(p)\subset S.$ 
Since $p\in X=\overline {B_r(x)}$ we may take some  $q\in B_b(p)\cap B_r(x).$  
Let $c>0$ be small enough that $B_c(q)\subset B_b(p)$ and $B_c(q)\subset B_r(x).$
Then $B_c(q)\subset B_b(p)\subset S$ and $B_c(q)\subset B_r(x)\subset X.$ Therefore  $B_c(q)\subset S\cap X=int_X(X\cap A_{j_0}).$
Addendum : More simply: In any space $M,$ if $S$ is open in $M$ and $X=\overline Y$, then $S\cap X\ne \phi \implies S\cap Y\ne  \phi.$..... So in the Q, with  $Y=B_r(X)$ we have $\phi\ne S\cap Y\subset S\cap X=int_X(X\cap A_{j_0}),$ and $S\cap Y$ is open in $M.$   
