If $|f(x) − f(y)| ≤ |\sin x − \sin y|,\text{for all} ~~x, y \in\mathbb{R}$ then $\text{Show that} ~f ~~\text{is}~~ 2\pi−\text{periodic}~?$

Question

Problem :

Let : $f : \mathbb{R} ~ \to ~\mathbb{R}~~\text{be such that}~: $

$$\color{orange}{|f(x) − f(y)| ≤ |\sin x − \sin y|,\text{for all} ~~x, y \in\mathbb{R}}$$

$$\color{red}{•~\text{Show that} ~~f ~~\text{is}~~ 2\pi−\text{periodic}~?}$$

$$\color{red}{•\text{Show that} ~f~ \text{is differentiable at }~\frac{\pi}{2}~\text{then compute}~ f'\left(\frac{\pi}{2}\right)~?}$$

I know that :

function $f$ is $2\pi−$periodic if $∀x\in\mathbb{R}~, f(x + 2\pi) = f(x)$ but how I applied here? Also second question how ? Can some one give me a hints or ideas?

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Edit

Thanks to Fred's for given a good solution but not complete now we need prove that there no $k≤2π$ such satisfy the equation $f(x+k)=f(x)$.

Answer 1

$|f(x+ 2 \pi)-f(x)| \le | \sin(x+2 \pi)- \sin x|=0.$ Conclusion ?

$|\frac{f(x)-f( \pi/2)}{x- \pi/2}| \le |\frac{\sin(x)-\sin ( \pi/2)}{x- \pi/2}| $

What is $ \lim_{x \to \pi/2}\frac{\sin(x)-\sin ( \pi/2)}{x- \pi/2}$ ?