 Solution

The Baire category theorem states that every countable intersection of open dense subsets of a complete metric space is dense. Equivalently, no nonempty open subset of a complete metric space is a countable union of nowhere dense sets.

To prove the first form, let $G_1,G_2,\dots$ be open dense subsets of a complete metric space $X$, and let $V\subseteq X$ be nonempty and open. Choose a closed ballInductively, density and openness of $G_{n+1}$ allow a closed ballThe balls are nested and $d(x_m,x_n)<2^{-n}$ for $m>n$, so $(x_n)$ is Cauchy. Completeness gives $x_n\to x\in X$. For every $n$, all later centres lie in $\overline{B}(x_n,r_n)$; closedness gives $x\in \overline{B}(x_n,r_n)\subseteq G_n$. The first ball also lies in $V$, henceSince every nonempty open $V$ meets the intersection, that intersection is dense.