Let CNC denote the statement that for any countable descending chain of non-empty sets:
$$A_0 \supseteq A_1 \supseteq A_2 \supseteq [...]$$
There is a choice function $f$ so that for all $n$, $f(A_n) \in A_n$.
This seems to be an interesting choice principle, because it is equivalent to the sequence definition of continuity in a metric space, so it seems to enable a basic amount of analysis. It is also sufficient to guarantee no infinite Dedekind finite sets exist*. I'm not aware of any common choice principle strictly weaker than this, so it seems like a minimal choice principle that can prevent a lot of very counterintuitve things from happening.
This is clearly weaker than countable choice, but I can't prove it isn't equivalent to countable choice. If it is, that would obviously answer my question. Though I suspect it isn't. I would very much like to see a proof of this either way.
* let $X$ be an infinite set. Then let $A_n$ be the set of finite ordered subsets of $X$ with cardinality $\ge n$. This is clearly a descending chain, and any choice function on the $A_n$ gives an infinite sequence of ordered subsets of $X$. The ordered subsets get arbitrarily large, so their union must be infinite. And we can use the ordering of the subsets and the fact they come in a sequence to wellorder their union. So their union must be equipotent to $\mathbb{N}$, so $X$ has a subset equipotent to $\mathbb{N}$.
It is not clearly weaker than countable choice, but it is clearly a consequence. The two are equivalent.
Given a countable family of non-empty sets, $\{X_n\mid n\in\omega\}$ let $T_n$ denote the product $\prod_{k<n}X_k$. Consider $A_n=\bigcup_{k>n}T_k$.
Clearly, $A_{n+1}\subseteq A_n$. Now, let $f$ be a choice function, then $f(n)\in T_k$ for some $k>n$, so $n$ is in the domain of $f(n)$. Let $g(n)=f(n)(n)$, then $g(n)\in X_n$ for all $n$, making $g$ a choice function.