Under Dependent Choice, a nonempty complete metric space is not a countable union of closed sets with empty interior

If $X$ is empty the assertion is vacuous. Otherwise choose an open ball $B_0$ whose closure lies in $O$; this is possible because $O$ is open.

Given a nonempty open ball $B_{n-1}$, its intersection with $X\setminus F_n$ is nonempty because $F_n$ has empty interior. Choose an open ball $B_n$ with nonempty closure, $\overline{B_n}\subseteq B_{n-1}\setminus F_n$, and radius below $2^{-n}$.

Dependent choice gives balls $B_n$ satisfying step 1.2 for every $n$. Choose centres $x_n\in B_n$.

The nesting and the radius bound make $(x_n)$ Cauchy: for $m>n$, both $x_m$ and $x_n$ lie in $\overline{B_n}$, so their distance is at most twice the radius of $B_n$, which tends to zero.

Let $x=\lim_nx_n$, supplied by completeness [L1]. For every $n$, the tail lies in the closed set $\overline{B_n}$, so $x\in\overline{B_n}\subseteq B_{n-1}\setminus F_n$.

Thus $x\in O\setminus\bigcup_nF_n$. Every nonempty open $O$ meets this complement, proving both stated formulations.