Suppose $f:\mathbb{R}\to\mathbb{R}$ is a continuous function and satisfies the "reverse-Lipschitz" criterion. That is, for all $x,y\in\mathbb{R}$ and some $k> 0$, we have \begin{equation*} \lvert f(x)-f(y) \rvert \geq k \lvert x-y \rvert \end{equation*} How do we go about showing that such a function is a surjection?
Breaking $\mathbb{R}$ into postive and negative values has provided me with inequalities \begin{equation} kx-\lvert f(0) \rvert \leq \lvert f(x) \rvert \text{ if $x \geq 0$} \end{equation} \begin{equation} -kx-\lvert f(0) \rvert \leq \lvert f(x) \rvert \text{ if $x \leq 0$} \end{equation} Which both show $\lvert f(x) \rvert$ is unbounded, and if I could show in general that one set of values has no upper bound and the other no lower bound, then I could finish instantly by applying the Intermediate Value Theorem I believe.
Alternatively, I've considered trying to show that the image of $f$ is both open and closed and thus must be all of $\mathbb{R}$ but have made as equally little progress there.