= Solution
A subset $A$ of a metric space $X$ is a <nowhere dense set> when
$$
\operatorname{int}\overline A=\varnothing.
$$
Equivalently, every nonempty open set contains a nonempty open subset disjoint from $A$.
The <Baire category theorem> says that if $X$ is a complete metric space and $G_1,G_2,\ldots$ are open dense subsets of $X$, then $\bigcap_{n\geq1}G_n$ is dense in $X$. Equivalently, no nonempty open subset of $X$ is a countable union of nowhere-dense sets.
To prove it, take a nonempty open set $V$. Since $G_1$ is open and dense, there is a closed ball
$$
\overline B(x_1,r_1)\subset V\cap G_1
$$
with $0<r_1<2^{-1}$. Inductively, openness and density of $G_{n+1}$ allow us to choose
$$
\overline B(x_{n+1},r_{n+1})
\subset B(x_n,r_n)\cap G_{n+1},
\qquad 0<r_{n+1}<2^{-(n+1)}.
$$
The balls are nested, and for $m>n$ their centres satisfy
$$
d(x_m,x_n)<r_n<2^{-n}.
$$
Thus $(x_n)$ is Cauchy and has a limit $x\in X$. Every closed ball contains the tail of the sequence, so it contains $x$. Hence
$$
x\in V\cap\bigcap_{n\geq1}G_n.
$$
As $V$ was arbitrary, the intersection is dense.
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