Let $f:[0,1]\to \mathbb R$ s.t. there is $a\in (0,1)$ s.t. $$\forall x,y\in [0,1], |x-y|\geq a\implies |f(x)-f(y)|<|x-y|.$$
Show there is $C\in [0,1)$ s.t. $$\forall x,y\in [0,1], |x-y|\geq a\implies |f(x)-f(x)|\leq C|x-y|.$$
As Hint I have that : Set $E=\{x,y\in [0,1]\mid |x-y|\geq a\}$. Prove that $$F:E\to \mathbb R, (x,y)\longmapsto \left|\frac{f(x)-f(y)}{x-y}\right|,$$ is continuous. I really don't understand why $F$ is continuous... so
Q1) Why is $F$ continuous ?
Attempt
I tried to do an other way. Suppose that for all $n$, there is $x_n,y_n\in [0,1]$ s.t. $$|x_n-y_n|\geq a\quad \text{and}\quad |f(x_n)-f(x_n)|\geq \left(1-\frac{1}{n}\right)|x_n-y_n|.$$
I suppose WLOG that $(x_n)$ and $(y_n)$ converge to $x,y\in [0,1]$. Then $|x-y|\geq a$ and since $$|x_n-y_n|>|f(x_n)-f(y_n)|\geq \left(1-\frac{1}{n}\right)|x_n-y_n|,$$ we get $$\lim_{n\to \infty }|f(x_n)-f(y_n)|=|x-y|\geq a,$$ but I don't see in what it's a contradiction.