Strength of "Cofinite Choice" Let $A$ be a set of sets such that $(\bigcup A) \setminus x$ is finite for every $x \in A$.
How strong is the variant of AC asserting that such an $A$ has a choice function?
I'd assume it to be weaker than countable choice, but ZF doesn't seem to prove it either.
 A: The axiom of cofinite choice is equivalent to the principle that every Dedekind-finite set is finite. Equivalently, for every infinite set $X$, there is an injective function $\omega\to X$. So as you suspected, it is (strictly) weaker than countable choice, but not provable in ZF.
Assume cofinite choice and let $X$ be an infinite set. Let $A$ be the set of all cofinite subsets of $X$, and let $g$ be a choice function for $A$. Then we can construct an injective function $\omega\to X$ by recursion. Let $f_0\colon 0\to X$ be the empty function. Given an injective function $f_n\colon n\to X$, the set $X_n = X\setminus \text{im}(f_n)$ is cofinite, and we can extend $f_n$ to $f_{n+1}\colon (n+1)\to X$ by defining $f_{n+1}(n) = g(X_n)$. Then $f = \bigcup_{n\in \omega} f_n$ is an injective function $\omega\to X$.
Conversely, assume that every Dedekind-finite set is finite. Let $A$ be a set of nonempty sets such that $(\bigcup A)\setminus x$ is finite for all $x\in A$. If $\bigcup A$ is finite, then $A$ is finite, and ZF proves that every finite set has a choice function. If $\bigcup A$ is infinite, let $f\colon \omega\to \bigcup A$ be an injective function. Then for any $x\in A$, since $(\bigcup A)\setminus x$ is finite, $x\cap \text{im}(f)$ is nonempty. So we can define a choice function $g$ on $A$ by defining $g(x) = f(n)$ for the minimal $n\in \omega$ such that $f(n)\in x$.
