Prove that f is differentiable in $\mathbb{R}$

You have $|f(x)-f(y)|\le |x-y||x-y|$ then $\displaystyle\frac{|f(x)-f(y)|}{|x-y|}\le|x-y|$ Just take the limit as $x\to y$ , you get $|f'(y)|\le 0$(then $f'(y)=0)$ for all $y$ then $f$ is constant

I would like to point out that it is possible to prove the assertion without proving differentiability in the first place.

For each x we have

$$|f(x)-f(0)| = \left|\sum_{k=1}^n f\left(\frac{k}{n} \cdot x\right)-f\left(\frac{k-1}{n}\cdot x\right)\right|$$

Thus by the triangle inequality

$$|f(x)-f(0)| \le \sum_{k=1}^n \left|f\left(\frac{k}{n}\cdot x\right)-f\left(\frac{k-1}{n}\cdot x\right)\right|$$

Since for each $k$

$$\left|f\left(\frac{k}{n}\cdot x\right)-f\left(\frac{k-1}{n}\cdot x\right)\right| \le \left|\frac{k}{n}\cdot x-\frac{k-1}{n}\cdot x\right|^2 = \left|\frac{x}{n}\right|^2$$

It follows that

$$|f(x)-f(0)| \le \sum_{k=1}^n \frac{x^2}{n^2} = \frac{x^2}{n}$$

And thus

$$|f(x)-f(0)|\le \limsup_{n\rightarrow\infty} \frac{x^2}{n} = 0$$

Hence for all $x$

$$f(x)=f(0)$$

Thus f is constant. It follows that f is differentiable everywhere ;)

For each $x\in\mathbb R$, for each $y\in\mathbb R$ with $y\neq x$, we have $0\leq |\frac{f(x)-f(y)}{x-y}|\leq |x-y|$. Let $y\to x$. Then by squeezing we obtain $f'(x)=0$ for all $x\in\mathbb R$ so that $f$ is constant.