If $F: \mathbb{R}^{m} \rightarrow \mathbb{R}^{m}$ is continuous and $\| F(x) - F(y)\| \geq \lambda \| x - y \|$ is $F$ a surjection? In my real analysis class my professor gave us the problem of proving that if $F: \mathbb{R}^{m} \rightarrow \mathbb{R}^{m}$ is continuous and satisfies $\| F(x) - F(y)\| \geq \lambda \| x - y \|$ then $F$ is a bijection with continuous inverse. ($∥⋅∥$ is the Eucliden norm and $\lambda$ is some positive real number.)
The problem of injectivity is easy enough since if $x \neq y$ then $\|F(x) - F(y)\| \geq \lambda \|x-y\| > 0$.
Also given that F is a continuous bijection then the continuity of the  inverse $g$ is also obvious since fixing $x = F(u)$ and $y = F(v)$ we have that $\|g(x) - g(y)\| = \|u-v\| \leq \frac{1}{\lambda}\|x-y\|$  so g is Lipschitz and therefore continuous.
My question is, how exactly is one supposed to prove surjectivity? It seems easy enough by intermediate value theorem if we restrict $F:\mathbb{R} \rightarrow \mathbb{R}$. But I can't seem to figure it out more generally. any hints would be much appreciated!
 A: This is a partial answer.
Call $X=F(\mathbb{R}^m)$. Then the inverse map $g: X \longrightarrow \mathbb{R}^m$ is Lipschitz continuous. I show that this implies that $X$ is a closed subset of $\mathbb{R}^m$.
Let $x_n \subset X$ be a convergent sequence with $x_n \to x$. It is enough to show that $x \in X$. Now, $g(x_n)$ is a Cauchy sequence in $\mathbb{R}^m$, so it converges to some $y$.  Applying $F$ we get $x_n \to F(y) $ hence $x = F(y) \in X$.
If one is able to show that $X$ is open as well (maybe using that $F$ is an embedding?), by connectedness argument we have $X = \mathbb{R}^m$.
A: here's an idea with $m=2.$ 
First take $\lambda=1.$ We can rescale to reduce to this case.
Now take the axes, these must map to curves going off to infinity. So we have a curly X in the image. 
Let's take a point $(a,b)$ and try to find the point mapping to it. It must be in one part of the $X.$ 
Let's assume WLOG the part is the image of the upper right quadrant.
If we go far enough along the axes to points $(x,0)$ and $(0,y)$ the line between must map to a curve giving a curly triangle than contains $(a,b)$. This follows from the main hypothesis since the mapped to point must be further from the origin.
Now consider the rays going from the origin to the line between $(x,0)$ and $(0,y).$ We can parametrize these by distance from $(x,0).$ At some point their images must transit from being below $(a,b)$ to above it. So take the supremum of points which give curves that lie below. $(a,b)$ must lie on the image of that ray. 
