TITLE: proof of the surjectivity of a function that satisfies certain properties
QUESTION [7 upvotes]: Let $f\colon \mathbb R^2 \to \mathbb R^2$ be a continuous function such that $|f(p)-f(q)|\geq a|p-q|,\quad \forall p,q\in\mathbb R^2$ and $a>0$. Show that $f$ is injective and surjective (therefore has inverse) and that its inverse is continuous.
This is a problem from a metric space topology test that I did. The most important contents of the test were uniform convergence, equicontinuity, Arzelà-Ascoli Theorem, iterated functions, Stone-Weierstrass Theorem etc. The exercise is very simple and I believe it is possible to solve it by more elementary concepts.
I already showed that $f$ is injective so my problem is surjectivity. For surjectivity i've tried something like this: since $f$ is injective it follows that $\exists g \colon \mathbb R^2 \to \mathbb R^2$ such that $\left(g\circ f\right)(x)=x$. On the other hand $f$ is a right inverse for $g$ which implies that $g$ is surjective. I was trying to show that $g$ is injective but i did not get anything. 
I have also shown that $f^{-1}$ is continuous (assuming $ f $ is surjective).
Thank you very much!

REPLY [4 votes]: For first think you can observe that the map is closed because if you consider $C\in\mathbb{R}^2$ closed you have that if $y\in\mathbb{R}^2$ is  in the clousure of $f(C)$ than there exists $\{x_n\}_n\subset C$ for which $f(x_n)\to y$. So $\{f(x_n)\}_n$ is a Cauchy sequence and by your property of $f$ you have that also $\{x_n\}_n$ is a Cauchy sequence and so it is convergent to an $x\in C$ because $C$ is closed. But $f$ is continuos and so  $f(x_n)\to (f(x)=y)$. So $y\in f(C)$ and than $f(C) $ is closed.
Now my idea is that if you prove that $f$ is  also an open map than it is surjective because each open-closed continuos map from a topological space  to a connected topological space is surjective.
Another idea is to prove that the image of the function is dense in $\mathbb{R}^2$ because in this case
$\mathbb{R}^2 =cl(f(\mathbb{R}^2))=f( \mathbb{R}^2)$