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.
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 $f(n)\in x$.
