In ZFC I want to prove the following result:
Proposition 1: Let $A$ be a set and let ${(G_k)}_{k \in \mathbb N}$ be a (countable) family of countable nonempty subsets of $A$. Then there exist a mapping $f: \mathbb N \to A$ satisfying:
$\tag 1 \text{There exist a partition of } \mathbb N \text{ into a family of subsets } N_k , \, k \ge 0 $
$\tag 2 f \text{ restricted to } N_k \text{ is an injective mapping onto } \,G_k, \, k \ge 0$
We can see immediately that $f$ maps onto $\bigcup G_k$ and that $f$ is injective when the $G_k$ are mutually disjoint.
My hand-waving idea of a proof:
Using the axiom of choice, we can endow all the $G_k$ with a well-ordering relation. The idea is is to define, using recursion, a way of visiting each index $k$ for the $G_k$ family as many times as necessary to 'scratch off and exhaust' the elements in $G_k$ as they are enumerated. For instance, if $G_k$ is finite with $\alpha_k$ elements, we would 'visit' $k$ exactly $\alpha_k$ times.
Another idea:
Proposition 1 is sure looks like it is equivalent to showing that a countable union of countable sets in countable; the arguments found in this math.stackexchange question make this very plausible.
Question 1:
If proposition 1 is valid in ZFC, any ideas on how to go about proving it.
Question 2:
Assuming again that proposition 1 is true, is it equivalent in ZF (axiom of choice dropped) to the Axiom of Choice from a 'Countable Family of Countable Sets':
Axiom AOC.CFCS: If ${\displaystyle (S_{i })_{\,i \in \mathbb N}}$ is a family of non-empty countable sets indexed by the natural numbers, then $\;{\displaystyle \prod _{i \in \mathbb N}S_{i }\neq \emptyset }$.
A 'yes' or 'no' would be helpful here with perhaps some idea sketch/links to help me see this.