Cardinality of an infinite union of finite sets

If the axiom of choice is assumed then we can simply well order everything and just define by induction an injective map.

Simply choose $f_i\colon E_i\hookrightarrow\mathbb N$ which is an injection. Since we assume the axiom of choice we also have:

$$\tag{1}|I|=|I\times I|\ge|I\times\mathbb N|=|I|$$ We can now inject $E_i$ into $\{i\}\times\mathbb N$, and use $(1)$ to obtain a bijection into $|I|$, and thus into $I$.

Without the axiom of choice it is consistent to have a countable union of disjoint pairs, and the result is not at all countable, so there is no such injection.

(This example is of course Russell's well known saying that you need the axiom of choice to choose from infinitely many pairs of socks.)