Let $f:\Bbb R\rightarrow\Bbb R$ be such that $|f(x)-f(y)|\le M|x-y|^\alpha$. Prove that $f$ is constant.

Since $x\neq y\Longrightarrow\left|\frac{f(y)-f(x)}{y-x}\right|\leqslant M|x-y|^{\alpha-1}$ and since $\alpha-1>0$, $(\forall x\in\mathbb{R}):f'(x)=0.$ Therefore, $f$ is constant.

Let $a $ be an arbitrary real.

for $x\ne a $, $$|f (x)-f (a)|\le M|x-a|^\alpha $$ and

$$\frac {|f (x)-f (a)|}{|x-a|}\le M |x-a|^\beta $$ with $$\beta =\alpha-1>0$$

thus if $x $ goes to $a ,$

$f (x)-f (a) $ goes to zero and

$\frac {f (x)-f (a)}{x-a} $ goes to zero.

this means that $f $ is continuous at $a $ and differentiable at $a $ with $$f'(a)=0$$

You can finish.