TITLE: What's "the catch" in this question?
QUESTION [9 upvotes]: I am solving old exam questions and I came across this question:

Let $\langle A_n \mid n < \omega\rangle$ disjoint sets such that $\bigcup_{n < \omega}A_n = \mathbb{R}$. Prove that there exists $n < \omega$ such that $|A_n| = |\mathbb{R}|$

It seems too easy - if the statement is false, then the cardinality of $\bigcup_{n < \omega}A_n$ is at most $\aleph_0$, which is less than $2^{\aleph_0}$, which is the cardinality of $\mathbb{R}$.
What am I missing?
Thanks!

REPLY [13 votes]: 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.