DC and finite multiple selections

Under DC take an $R$-path and put $F_n=\{x_n\}$. These are DMC levels. Countable finite choice follows from DC as well.

Conversely, take DMC levels for a serial relation. For each $n$ the set of linear orders on the nonempty finite $F_n$ is nonempty and finite (enumerate that single finite set to see this). Countable finite choice supplies an order ${<}_n$ for every $n$.

Under MC select, once for all $x\in X$, a finite nonempty $H(x)\subseteq R[x]$. Fix $a\in X$ and set $F_0=\{a\}$, $F_{n+1}={\bigcup }_{x\in F_n}H(x)$. A finite union of finite sets is finite by finite induction, and each member has a successor in the next nonempty level. This recursion proves DMC.

Start at the ${<}_0$-least point and take the ${<}_{n+1}$-least $R$-successor in $F_{n+1}$. Such a successor exists by the universal successor clause of DMC. The rule is a self-map on tagged states $(n,x)$, so recursion supplies a path. Starting-point-free DC now implies full DC.

Under countable choice select $x_n\in X_n$ and use $\{x_n\}$ for the CMC selection.