@JonasTeuwen I have a new one. This is a $**$ in Spiavk's, meaning it is extra hard. I know you like when stuff is extra hard.
Suppose $f$ is continuous in $[0,1]$ and $f(0)=f(1)$. Let $n$ be any natural number. Prove there exists an $x$ such that $f(x)=f(x+1/n)$
I think I need to exploit the fact that for continuous $f$ $$\lim_{\epsilon \to0}f(x+\epsilon)=f(x)$$
If $\forall n\forall x$ $$g(x)=f(x)-f(x+1/n)$$ is $\neq0$, $f$ would not be continuous.
This stems for $\forall \epsilon >0 \exists n\in\Bbb N$ such that $$\frac 1 n <\epsilon$$.
@BrianM.Scott I'm on the right track, yes?