Proof without ordinals that $\left|\bigcup_{i\in I}C_i\right|\leqslant |I|$? Let $I$ be an infinite set and each $C_i$ be a nonempty countable (finite or infinite) set.
I would like to prove without ordinals/cardinal arithmetic that 
$$\left|\bigcup_{i\in I}C_i\right|\leqslant |I|,$$
where $|\cdot|\leqslant|\cdot|$ is thought of only as an order relation based on existence of an injective function. 
I have seen this fact justified with cardinal arithmetic, as at the end of this answer. However, I would like to be able to prove this with minimal set-theoretic background, if possible. All proofs I have found justifying cardinal arithmetic, such as the rule $\aleph_\alpha\cdot\aleph_\beta=\aleph_{\max(\alpha,\beta)}$, rely on ordinals. Thus I would rather avoid taking any rules of cardinal arithmetic as given and just show the existence of an injective function $f:\cup_{i\in I}C_i\to I$ directly.
It is fine if the proof uses axiom of choice or Zorn's lemma, as I imagine this is necessary.
 A: Okay, we are going to do a sequence of reductions, and then I will send you elsewhere for the final details.


*

*It is enough to prove that $|I\times\Bbb N|=|I|$, if this is true, then we can choose for each $i$, an injection from $C_i$ into $\Bbb N$, and for each $x\in\bigcup C_i$, choose $i$ such that $x\in C_i$ (there may be several, you see), then combine these choices into an injection from $\bigcup C_i$ into $I\times\Bbb N$.

*It is enough, therefore, to show that there is some set $J$ such that $|I|=|J\times\Bbb N|$. Indeed, if this is true, then $$|I\times\Bbb N|=|(J\times\Bbb{N)\times N}|=|J\times(\Bbb{N\times N})|=|J\times\Bbb N|=|I|.$$

*Now we can use Zorn's lemma as in Prove that if $A$ is an infinite set then $A \times 2$ is equipotent to $A$ or Show that an infinite set $C$ is equipotent to its cartesian product $C\times C$ (the latter proves an even stronger statement, but goes through a similar step), or The cardinality of an uncountable set times $\mathbb N$..
Good luck.
