What's "the catch" in this question?

Solution 1:

You can't assume that all the sets are countable. I mean, $\aleph_\omega$ is uncountable and can be written as a countable union of smaller sets.

The point is to use König's theorem, and prove that can't be true. That is, $\operatorname{cf}(2^{\aleph_0})>\aleph_0$.

As Arthur points out, this is false without the axiom of choice, as it is consistent that the real numbers are a countable union of countable sets.

Solution 2:

What you are missing is that the cardinal number $|\mathbb R|=2^{\aleph_0}$ is not necessarily equal to $\aleph_1$. How do you know that $2^{\aleph_0}$ is not equal to $\aleph_\omega$? It's not, but that's the point of that old exam question. The assumption that $2^{\aleph_0}=\aleph_1$ is called the continuum hypothesis.