Here's an interesting question I stumbled upon recently: can a Lipschitz-continuous function $f:[0,1]\to[0,1]$ with Lipschitz bound $L$ take on every value $y\in[0,1]$ more than $L$ times?
Intuitively, I'd expect the answer to be no, but it's not at all obvious to me how that could be proven. It seems to me like there might necessarily be some measure theory involved, but I'm curious whether a more elementary proof exists as well.
Here is a proof using quite a big gun, namely the Coarea formula. This formula shows that $$ \int_{(0,1)} |f'(x)| \, dx = \int_{\mathbb{R}} \# f^{-1}(\{t\}) \, d t, $$ where $f'(x)$ is the derivative of $f$, which exists almost everywhere by Rademacher's theorem and which satisfies $|f'(x)| \leq L$ (whenever it exists) if $f$ is $L$-Lipschitz. In the above formula, we have used that the $0$-dimensional Hausdorff measure is just the counting measure.
Now, first note that the left-hand side satisfies $\int_{(0,1)} |f'(x)| \, dx \leq L$.
But if $f$ takes on each $y \in [0,1]$ strictly more than $L$ times, then this means $\# f^{-1}(\{t\}) > L$ for all $t \in [0,1]$, so that the right-hand side of the equation satisfies $$ \int_{\mathbb{R}} \# f^{-1}(\{t\}) \, d t \geq \int_{[0,1]} \# f^{-1}(\{t\}) \, d t > L. $$ In combination, we get the desired contradiction.