Let $f:[0,1]\to \mathbb{R} $, a continuous function and differentiable, that verifie:
$|f(x)-f(y)|<=|x-y| \forall x,y\in [0,1]$
Prove that $\forall x,y\in [0,1] :|f(x)-f(y)|\leq1/2$.
I began with a reasoning by contradiction, Suppose that $\exists \alpha,\beta \in [0,1]: |f(\alpha)-f(\beta)|>1/2$
Using the hypothesis we can proof easily that f is uniformaly continuous,so: $\forall\epsilon>0 , \exists\delta>0,\forall x,y\in [0,1]: |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon$
In particular $\epsilon=1/2$
$\exists\delta>0,\forall x,y\in [0,1]: |x-y|<\delta \Rightarrow |f(x)-f(y)|<1/2$
My problem is how to take all $x$ and $y$ whitout $|x-y|<\delta$ to reach th econtradiction and I don't know how to use the differentiability of $f$.
PS: I thought about "Mean value theorem" but I don't see any contradiction, with manipultations I found that it must be $|x-y|>1/2$.
Take $$f(x)=\frac 34 x$$
then for $(x,y)\in[0,1]$,
$$f(x)-f(y)=\frac 34 (x-y)$$
and
$$|f(x)-f(y)|\le |x-y|$$ but $$f(1)-f(0)=\frac 34>\frac 12$$