AC implies DC implies countable choice

Assume AC, let $R$ be serial on $X\mathrm{\ne }\mathrm{\varnothing }$, and fix $a\in X$. Apply AC to the successor sets $R[x]=\{y:xRy\}$ to get a function $s:X\to X$ with $xRs(x)$. Repeated successor sets use the same selected value.

Assume DC and let $(X_n)$ be a nonempty-set family. The set $T$ of finite functions $t$ with domain some $n<\omega$ and $t(i)\in X_i$ contains the empty function. One-step extension is serial: for a particular $t$, one point of $X_n$ extends it. DC starting at the empty function gives nested $t_n$ of domain $n$. Their union is a function on omega selecting from each $X_n$.

Recurse with $x_0=a$ and $x_{n+1}=s(x_n)$. This is the prescribed path and proves AC implies DC.

The remaining implications restrict the eligible input families: nonempty finite sets are nonempty sets, and pairs are finite nonempty sets; AC also applies to any set-indexed family of pairs. This includes singleton-valued families without additional choices.