Countable Union of Countable Sets 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? 
 A: Suppose $\{A_i\}$ are a countable family of countable sets and $\{B_i\}$ partition the integers into countably many disjoint countably infinite subsets.  You're right that the construction of $B_i$ doesn't necessarily use choice.  But let's talk about the truly important map - the injection from $\cup A_i$ into $\cup B_i$, which will show that $\cup A_i$ is countable.
Assume the $A_i$ are disjoint for simplicity.  How can we create this injection?
Start by injecting $A_1$ into $B_1$, which is possible since both are countable.  Well, there are a bunch of ways of doing this.  We need to pick one.  There's no canonical/best choice.  Call it $f_1$.  Then inject $A_2$ into $B_2$.  We have to pick another injection $f_2$ for this.  Then the union of the maps $f_i$ gives us the injection that we want.  But despite the fact that each $f_i$ is individually guaranteed to exist, we still have to pick a particular injection for each index $i$.  This requires the axiom of countable choice (more precisely, you need an element from the infinite product of the sets $\{g: g$ is an injection from $A_i$ to $B_i$$\}$).
A: Given a partition N into pairwise disjoint infinite subsets $\{S_j:j\in N\}$ and  given a family $\{T_j:j\in N\}$ of countable sets, there exists, for each $j\in N,$ an injective map  $f_j:T_j \to S_j.$ (And if $S_j\ne \phi$ there are many such maps.)But then you want to assert the existence  of  a sequence $(f_j)_{j\in N}$ of such maps. This requires  Countable Choice . Consider, for each $j\in N,$ the set $I_j$ of injections from $T_j$ into $S_j.$ We have $\forall j\in N \;(I_j\ne \phi).$ But you need  a Choice-function $\psi:N\to \cup_{j\in N}I_j$ such that $\forall j\in N\;(\psi (j)\in I_j).$  Then can you obtain a sequence $(f_j)_{j\in N}=(\psi (j))_{j\in N}$. Conversely, given the desired sequence $(f_j)_{j\in N},$ you obtain a Choice-function $\psi$ by defining $\psi (j)=f_j$ for all $j\in N.$
