Continuous function that verifies $|f(x)-f(y)|\ge|x-y|$

Question

Let $f: [0,1]\to\Bbb{R}$ be a continuous function verifying $f(0)=0$, $f(1)=1$ and $|f(x)-f(y)|\ge|x-y|$ for all $x,y\in[0,1]$. I have to prove $f(x)=x$ for all $x\in[0,1]$, but I honestly don't know even how to start.

Answer 1

The condition implies that $f$ is injective: If $f(x)=f(y)$, then $|x-y|\le 0$. Hence $f$ is monotonic. Let $0<x<1$. Then $0<f(x)<1$ and the inequality gives us $f(x)-0\ge x-0$ and $1-f(x)\ge 1-x$, hence $f(x)=x$.