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Why can't I use this proof to prove that the countable union of countable sets is countable without the axiom of countable choice?
Take the set of integers; some proper subset of it, call it $A$, can be mapped, with $f$, a 1-1 and onto function, to the set of even integers.
Some proper subset of the set of even integers, call it $B$, can be mapped, with $g$, a 1-1 and onto function, to the integers. Take the subset of $A$, call it $C$, which is mapped, with the use of $f$, to $B$ which is mapped, with the use of $g$, to the set of integers. Then find the subset of $A$ which is mapped, in a similar fashion, to $C$. Continue doing this forever.
Each one of these sets is countable, given that that they map to $A$ in a 1-1 and onto fashion.
We can find a countable amount of them (just keep doing the same thing). They are all subsets of $A$ and so their union is a subset of $A$.
One can therefore easily show, using the Cantor Bernstein Theorem, that they have the same carnality as $A$. We can easily show that every union of countable sets can easily be placed in a 1-1 and onto fashion with this specific union of countable set.
I assume that I never used the Axiom of Countable Choice in this proof, so what is the problem?