Serial Dependent Choice implies the complete-metric Baire principle over ZF

If $X=\mathrm{\varnothing }$, the intersection is empty and dense. Otherwise it suffices to meet an arbitrary nonempty open $V\subseteq X$. Set $C(c,r)=\bar{B}(c,r)$. For any nonempty open $W$, fix $c\in W$ and $\delta >0$ with $B(c,\delta )\subseteq W$. Given a bound $b>0$, take $t\ge 1$ with $r=1\mathrm{/}t<\min (\delta ,b)$. Then $r$ is positive rational, $r\le b$, and $C(c,r)\subseteq B(c,\delta )\subseteq W$, directly from $d(c,x)\le r<\delta$.

By ZF Separation, let $S$ consist of all triples $(n,c,r)\in \omega \times X\times \mathbb{Q}$ with $0<r\le 1\mathrm{/}(n+1)$ and $C(c,r)\subseteq V\cap U_n$. The set $V\cap U_0$ is nonempty by density and open. The preceding construction with $b=1$ gives an initial state $s_0=(0,c_0,r_0)\in S$.

Relate $(n,c,r)$ to $(n+1,c^{\mathrm{\prime }},r^{\mathrm{\prime }})$ when both lie in $S$ and $C(c^{\mathrm{\prime }},r^{\mathrm{\prime }})\subseteq B(c,r)\cap U_{n+1}$. For each state, $c\in B(c,r)$, so density of $U_{n+1}$ makes $W=B(c,r)\cap U_{n+1}$ nonempty; it is open. The construction with $b=1\mathrm{/}(n+2)$ supplies $c^{\mathrm{\prime }},r^{\mathrm{\prime }}$. Since $B(c,r)\subseteq C(c,r)\subseteq V$, this triple belongs to $S$. Thus the displayed relation is serial on the nonempty set $S$. Only one centre and radius were fixed for this one existence assertion.

Apply prescribed-start DC to $S$ with initial state $s_0$. The resulting chain has stage coordinate $n$ at position $n$: this holds at zero, and each relation step increments that coordinate by one. Write its states $(n,c_n,r_n)$ and $C_n=C(c_n,r_n)$. Then $C_{n+1}\subseteq B(c_n,r_n)\subseteq C_n$, $r_n\le 1\mathrm{/}(n+1)$, and $C_n\subseteq V\cap U_n$. This application is the proof's sequence-selection use of DC; centres are already components of the selected states.

For fixed $n$, induction on $m-n$ gives $C_m\subseteq C_n$ for all $m\ge n$: equality is the base, and the next containment follows from nesting. Since $c_m\in C_m$, for $m,k\ge n$ the triangle inequality gives $d(c_m,c_k)\le d(c_m,c_n)+d(c_n,c_k)\le 2r_n\le 2\mathrm{/}(n+1)$. For any positive rational $\epsilon$, choose $t\ge 1$ with $1\mathrm{/}t<\epsilon \mathrm{/}2$ and take $n=t$; then the displayed bound is less than $\epsilon$. Hence $(c_n)$ is Cauchy. Completeness supplies a single limit $x\in X$.

Fix $n$. If $d(x,c_n)>r_n$, put $\eta =d(x,c_n)-r_n>0$ and fix a positive reciprocal $e<\eta$. Convergence gives $m\ge n$ with $d(x,c_m)<e$. The tail bound and triangle inequality yield $d(x,c_n)\le d(x,c_m)+d(c_m,c_n)<e+r_n<\eta +r_n=d(x,c_n)$, which is impossible. Therefore $d(x,c_n)\le r_n$ and $x\in C_n$. In particular equality on a closed-ball boundary is allowed.

Thus $x\in V\cap {\bigcap }_n{U}_n$. Since $V$ was an arbitrary nonempty open set, the intersection is dense. This proves CM-Baire, and hence also its equivalent category formulations.