Let $g:\mathbb{R}^2\rightarrow\mathbb{R}^2$ such that $\forall x,y\in \mathbb{R}^2, \|x-y\|=1\Rightarrow \|g(x)-g(y)\|=1$. Prove that $g$ is an isometry.
I managed to prove the following: Let $A$ be a dense subset of $\mathbb{R}_+$, $g:\mathbb{R}^2\rightarrow\mathbb{R}^2$ such that $\forall x,y\in \mathbb{R}^2, \|x-y\|\in A\Rightarrow \|g(x)-g(y)\|=\|x-y\|$. Then $g$ is an isometry. I don't see how it helps to solve the problem, though.