Question
Let $f:\;\mathbb{R}\to\mathbb{N}$ and let $X_n$ be the set of reals mapped to the integer $n$. Show that for some $n,\;X_n$ has cardinality of the continuum.
This is straightforward if we use König's theorem: assuming the contrary, i.e. $\left|X_n\right|<2^{\aleph_0}$ for all $n$:
$$2^{\aleph_0}=\left|\bigcup_{\mathbb{N}} X_n\right|=\sum_{\mathbb{N}}\big|X_n\big|\overset{\text{K.T.}}<\prod_{\mathbb{N}}2^{\aleph_0}=\big(2^{\aleph_0}\big)^{\aleph_0}=2^{\aleph_0}$$ a contradiction.
This feels overkill, but perhaps it is not. Am I missing a more elementary solution to this problem?
Yes, you need Koenig's theorem, or at least the consequence that $\operatorname{cf}(\kappa)<\operatorname{cf}(2^\kappa)$. If $\operatorname{cf}(\frak c)=\omega$, then there would be a partition of the real numbers into countably many sets all of which have cardinality strictly less than the continuum. But Koenig's theorem is exactly what guarantees that this cannot happen.
Interestingly, this depends on the axiom of choice, and it is consistent that the axiom of choice fails and $\Bbb R$ is a countable union of countable sets. Or the union of two sets which have strictly smaller cardinality!