Countable intersection of dense open subsets in complete metric space 
Let $X$ be a complete metric space. Let $E_n$ be a nowhere dense subset of $X$ for every $n$. Let $M_n$ be a dense open subset of $X$ for every $n$.
Show that $\bigcap_n M_n$ is not contained in $\bigcup_n E_n$.

My attempt:
Since $X$ is a complete metric space, then $X$ is a Baire space. Hence $\bigcap_n M_n$ is dense in $X$ and $\bigcup_n E_n$ has empty interior.
However, this is not enough to show the statement. There must be something missing but I can't find out.

Suppose $\bigcap_nM_n\subset \bigcup_nE_n$. Let $U_n$ be the complementary subset of $E_n$, it is a dense open subset. The family of open subsets $U_m,M_n$ is numerable. Baire theorem implies that $\bigcap_{n,m}M_n\cap U_m$ is dense, but $\bigcap_{n,m}M_n\cap U_m\subset (\bigcup_nE_n)\cap_mU_m=\bigcup_n(E_n\cap_mU_m) $ is empty. Contradiction.
