Countable unions of countable sets It seems the axiom of choice is needed to prove, over ZF set theory, that a countable union of countable sets is countable. 
Suppose we don't assume any form of choice, and stick to ZF. What are the possible cardinalities of a countable union of countable sets? 
Could a countable union of countable sets have the same cardinality as that of the real numbers? 

Yes. It is consistent that the real numbers are a countable union of countable sets. For example in the Feferman-Levy model this is true. In such model $\omega_1$ is also the countable union of countable sets (and there are models in which $\omega_1$ is the countable union of countable sets, but the real numbers are not).
It is consistent [relative to large cardinals] that $\omega_1$ is the countable union of countable sets, and $\omega_2$ is the countable union of countably many sets of size $\aleph_1$, that is $\omega_2$ is the countable union of countable unions of countable sets.
Indeed it is consistent [relative to some large cardinal axioms] that every set can be generated by reiterating countable unions of countable sets.
For non-aleph cardinals the follow results holds:

Let $M$ be a transitive model of ZFC, there exists $M\subseteq N$ with the same cardinals as $M$ [read: initial ordinals] and the following statement is true in $N$:
For every $\alpha$ there exists a set $X$ such that $X$ is a countable union of countable sets, and $\mathcal P(X)$ can be partitioned into $\aleph_\alpha$ nonempty sets.
D. B. Morris, A model of ZF which cannot be extended to a model of ZFC without adding ordinals, Notices Am. Math. Soc. 17, 577.

Note that $\mathcal P(X)$ can only be mapped onto set many ordinals, so this ensures us that there is indeed a proper class of cardinalities which can be written as countable unions of countable sets.

Some positive results are:

*

*If $X$ can be expressed as the countable union of countable sets, and $X$ can be well-ordered then $|X|\leq\aleph_1$.


*If $\langle A_i,f_i\rangle$ is given such that $A_i$ is countable and $f_i\colon A_i\to\omega$ is an injection, then $\bigcup A_i$ is countable.


*The collection of finite subsets of a countable set, as well finite sequences are both countable, without the use of the axiom of choice (so there are still only a countable collection of algebraic numbers in $\mathbb R$).
