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!

I suppose that $X$ is the closed ball $\overline{B(x,r)}$, let $Y=\overline{B(x,r/2)}$, if $Y\cap A_{j_0}$ has a non empty interior considered as a subspace of the metric space $\overline{B(x,r/2)}$, it implies there exists an open  ball $B(x',r')$ of $Y$ (that you can contained also in the interior of $X$ since if $y\in B(x',r')\subset \overline{B(x,r/2)}, d(x,y)\leq r/2<r$ ) contained in the interior $ Y\cap A_{j_0}\subset B(x,r)\cap A_{j_0}$, but an open subset of the interior of  $X$ is an open subset of $M$ since it is the intersection $U\cap int(X)$ where $U$ is open in $M$.
