(Baire Category Theorem). Let $(X,d)$ be a (non-empty) complete space. Suppose $X=\bigcup_n F_n$ where each $F_n$ is closed. Then there exists $n$ such that $F_n$ has non-empty interior.
Proof
Suppose for the sake of a contradiction that for each $n$, $F_n$ has empty interior. In particular, that would imply there exists an element $x_1\in F_1^c$ (otherwise $X$ would be empty). So, we may obtain a closed neighborhood $G_1$ of $x_1$ such that $G_1\cap F_1=\varnothing$, where we can specify that $diamG_1<\frac{1}{2}$ (this exists because a singleton is such an example). Consider $F_1\cup F_2$. Since they have empty interior, we may find $x_2\in X\backslash (F_1\cup F_2)$, with a closed neighborhood $G$ of $x_2$ satisfying $diamG<\frac{1}{2^2}$, such that $G\cap (F_1\cup F_2)=\varnothing$. Set $G_2=G\cap G_1$. This set in particular, is a closed set containing $x_2$……
My question is, why is $G\cap G_1$ a closed neighborhood of $x_2$?
Best Answer
If you already know Baire, suppose that all $F_n$ have empty interior.
Define $D_n = X\setminus F_n$. Then $D_n$ is open (as $F_n$ is closed) and dense (as $\overline{X\setminus F_n}= X\setminus \operatorname{int}(F_n)=X$)
Then Baire's theorem says that $$\bigcap_n D_n = \bigcap (X\setminus F_n) = X\setminus (\bigcup_n F_n) = X\setminus X=\emptyset$$
would be dense in $X$, a contradiction. So the assumption that all $F_n$ have empty interior is incorrect. QED.