Suppose $f: \mathbb{R}^n \to \mathbb{R}^n$ is continuous and for some $\lambda>0$, we have $\|f(x)-f(y)\|\geq \lambda \|x-y\|$ for all $x,y\in\mathbb{R}^{n}$. Is $f$ surjective?
I can show the image is closed by showing it contains its limit points. So, if I can show it is also open, then I am done (I want to avoid using the Invariance of Domain theorem).
Obviously, $f$ is one-to-one. By considering the the inverse map defined from $E=f(\mathbb{R}^n)$ back to $\mathbb{R}^n$ we get the following equivalent formulation of the problem:
Suppose $E \subset \mathbb{R}^n$ is closed and $f\colon E \to \mathbb{R}^n$ is Lipschitz, one-to-one, and onto. Is $E$ necessarily equal to all of $\mathbb{R}^n$?