Tell me more ×  Let us denote by ACC the axiom of countable
choice, namely the assertion that the product of countably many
non-empty sets is non-empty, and denote by UCC the assertion
that a countable union of countable sets is countable.  UCC is a
simple theorem of ZF+ACC.  Proof Suppose for every $i\in\omega$
we have $X_i$ a countable set, and $X_i\cap X_j=\varnothing$ for
$j\neq i$.  Since $X_i$ is countable $O_i=\{f\colon
X_i\to\omega\mid f\ \text{ injective}\}$ is non-empty. We can
choose $f_i\in O_i$ by the axiom of countable choice, and
define: $$F\colon\bigcup X_i\to \omega\times\omega\colon\qquad
x\mapsto\langle n,f_n(x)\rangle$$ Where $n$ is the unique
$n\in\omega$ such that $x\in X_n$.  The Cantor pairing function
shows that $\omega\times\omega$ is countable and we are done.
Is the opposite assertion is true, namely ZF+UCC implies ACC? If
the answer is negative, does that imply at least some other
weaker form of choice?    • As noted by Emil Jeřábek below, UCC
implies the axiom of countable choic