Assume that $f : A \to A$ is a function such that $\forall x, y \in A,\|f(y)-f(x)\|=\|y-x\|$. Prove that $f(A)= A$

Question

Good morning, I'm trying to solve this problem:

My attempt:

Assume the contrary that there is $z \in A - f(A)$. Because, $f$ is Lipschitz continuous and $A$ is compact, $f(A)$ is compact. As such, $A-f(A)$ is open and thus there is $\mathbb B(z,r) \cap f(A) = \emptyset$.

After that, I'm stuck and unable to proceed. Could you please shed me some light? Thank you so much!