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 $x_0\in O$ and $r_0>0$ such that $\overline{B(x_0,r_0)}\subseteq O$; openness supplies a ball about $x_0$ in $O$, and a smaller positive radius has this property. Put $B_0=B(x_0,r_0)$.

Given $n\ge 1$ and a state $(n-1,x,r)$ with $r>0$, the open set $B(x,r)\setminus F_n$ is nonempty because $F_n$ is closed with empty interior. Choose $y$ in that set and $s>0$ small enough that $\overline{B(y,s)}\subseteq B(x,r)\setminus F_n$ and $s<2^{-n}$; openness and [L3] permit this for each individual state. Thus the relation from $(n-1,x,r)$ to $(n,y,s)$ with these properties is serial on the set of triples $(k,z,q)\in \mathbb{N}\times X\times (0,\mathrm{\infty })$.

Apply dependent choice to the serial state relation in step 1.2, starting at $(0,x_0,r_0)$. Its successive states have first coordinate $n$, so write them $(n,x_n,r_n)$ and put $B_n=B(x_n,r_n)$. The centres and radii are coordinate projections of this single DC sequence, not separately chosen from countably many balls. They satisfy $x_n\in B_n$, $r_n<2^{-n}$ for $n\ge 1$, and $\overline{B_n}\subseteq B_{n-1}\setminus F_n$.

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 $x={\lim }_n{x}_n$, 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 $x\in O\setminus {\bigcup }_n{F}_n$. Every nonempty open $O$ meets this complement, proving both stated formulations.