Let $I\subset \mathbb R$ be an interval, $f:I \to I$ continuous, and $x,y \in I$ such that $x<y$ and $f(x)=f(y)$. Show that for all $\varepsilon >0$ there are $a,b\in[x,y]$ such that $0<|a-b|\leq\varepsilon$ and $f(a)=f(b)$.
My proof (which is hopefully correct, see below) strikes me as unnecessarily clumsy. I wonder if one can prove it faster/more elegantly (e.g. by using stronger theorems).
Let $\varepsilon >0$. There exists $c\in \text{argmax}_{z\in[x,y]} |f(z)-f(x)|$. If $c\in\{x,y\}$ then the statement obviously holds. So assume $c\notin\{x,y\}$. Since $c\in (x,y)$ is a local maximum (WLOG) of $f$, there is $\delta>0$ such that for all $p\in [c-\delta, c+\delta]\cap[x,y]$ holds $f(p) \leq f(c)$. Now let $\kappa = \min\{\varepsilon/2, \delta, c-x, y-c \}$ and $a\in \text{argmax}\{f(c-\kappa), f(c+\kappa) \}$ and $b'\in \{c-\kappa,c+\kappa\}\setminus\{a\}$. Then $f(a)\in [f(b'),f(c)]$ and by the intermediate value theorem, there is $b$ between $b'$ and $c$ such that $f(b)=f(a)$.
I think this holds for all continuous functions on a compact interval.
Let $c$ be a point such that $f(c)$ is the maximum of $f$ in $[x,y]$. Note that since $f$ is continuous, it's injective iff it's strictly monotone. But for all $\epsilon > 0$, $f$ cannot be strictly monotone in $(c - \frac{\epsilon}{2}, c + \frac{\epsilon}{2})$ and hence it cannot be injective in this interval. This implies there are $a,b \in (c - \frac{\epsilon}{2}, c + \frac{\epsilon}{2})$ such that $f(a) = f(b)$ and $|a-b| < \epsilon$.