Surjectivity of real continuous expansive-type functions 
Let $f:\mathbb{R}^n \to \mathbb{R}^n$ be continuous and let there exist $\alpha > 0$ such that $||f(\mathbf{x}) - f(\mathbf{y})|| \geq \alpha || \mathbf{x} - \mathbf{y}||$ for all $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$. Prove that $f$ is one-one, onto and that $f^{-1}$ is continuous.

One-one is trivial. It is onto-ness that I can't show. 
Write $S = f(\mathbb{R}^n)$. Using sequential continuity, it is possible to show that $S$ is closed. If I could show $S$ is open, I would be done, but I can't.
Also, writing $g(\mathbf{x}) := \dfrac{f(\mathbf{x})}{\alpha}$, the condition can be converted to that of proper expansive map, $||g(\mathbf{x}) - g(\mathbf{y})|| \geq || \mathbf{x} - \mathbf{y}||$. But since $\mathbb{R}^n$ is not compact, I cannot use the result here.
Any help is appreciated!
EDIT: As commented below, the Invariance of Domain theorem seems to work in this case, but that result does not use the expansive-type condition provided here (except for showing the injectivity), and so it appears that an easier proof would be possible.
 A: Actually the Invariance of Domain theorem solves your problem, but not the way you think : indeed, if you apply it with $U = \mathbb{R}^n$, you only get that $f$ is a homeomorphism on its image, and you don't get that its image is $\mathbb{R}^n$. 
Here is the way you can use it correctly, and your expansion condition is really necessary :
For $r > 0$, denote by $B(a,r)$ the open ball of center $a \in \mathbb{R}^n$ and radius $r$. For all $r > 0$, the image $f(B(0,r))$ is open, by the Invariance of Domain theorem. So $f(B(0,r)) \cap B(f(0), \alpha r)$ is open in $B(f(0), \alpha r)$.
But $f(\overline{B(0,r)}) \cap B(f(0), \alpha r) = f(B(0,r)) \cap B(f(0), \alpha r)$ : indeed, if $||x|| = r$, then $||f(x)-f(0)|| \geq \alpha r$, so $f(x) \notin B(f(0), \alpha r)$. So you get that $f(B(0,r)) \cap B(f(0), \alpha r)$ is also closed in $B(f(0), \alpha r)$.
By connectivity, $f(B(0,r)) \cap B(f(0), \alpha r)$ is open and closed in $B(f(0), \alpha r)$, and not empty (because it contains $f(0)$), so it is equal to $B(f(0), \alpha r)$. In other words you have, for all $r > 0$, 
$$ B(f(0), \alpha r) \subset f(B(0,r)) $$
Obviously this implies that $f$ is surjective.
