Let $f: [-1, 1] \longrightarrow [-1, 1]$ such that $f\in C^{1}$. Prove that there's exist $x_{0} \in [-1, 1]$ such that $|f'(x_{0})| \leq 1$ 
Let $f: [-1, 1] \longrightarrow [-1, 1]$ such that $f$ is a class $C^{1}$ function. Prove that there's exist $x_{0} \in [-1, 1]$ such that $|f'(x_{0})| \leq 1$.

I know that $f'([- 1,1])$ is compact, since $f'$ is continuous. Therefore, it is closed and limited. To prove the result, I tried to use the continuity of $f'$ in some sequence and tried to use the Weierstrass theorem, but I could not conclude anything. I would like some suggestion.

Assume your claim is false. Your claim combined with the continuity of $f'$ implies that either $f'(x)>1$ is always true or that $f'(x)<-1$ is always true. Also, since $f'$ is continuous, we can take its integral: $$f(1)-f(-1) = \int_{-1}^1 f'(x)\,dx$$
If $f'(x)>1$ then $f(1)-f(-1)>2$. Similarly if $f'(x)<1$ then $f(1)-f(-1)<-2$. Using the triangle inequality, in both cases we have a contradiction.
