Lipschitz function, Berkeley problem 1.2.2

Problem 1.2.2
Suppose that $f$ maps compact interval $I$ into itself and that
$$mid f(x)-f(y) mid < mid x-y mid$$
for all $x, y in I, xneq y$. Can one conclude that there is some constant $M<1$ such that, for all $x, y in I$, $$mid f(x)-f(y) mid leq Mmid x-y mid$$?

If such function exists it is not continuously differentiable, otherwise $f’$ will attain supremum.

I tried to define $g: I times I rightarrow I$ s.t. $g(x,y)=frac{mid f(x)-f(y) mid}{mid x-y mid}$ which is continuous and attains supremum. But $g$ is not defined for $x=y$

Please give hint. Please do not give solution. Thanks!