How to show that every $\alpha$-Hölder function, with $\alpha>1$, is constant?

Derivatives are not needed here (which is a good thing since proofs based on them can lead to erroneous arguments, see the accepted answer for an example)...

Choose $y\gt x$ in the interval $(a,b)$. For every $n\geqslant1$, divide the interval $(x,y)$ into $n$ subintervals $(x_i,x_{i+1})$ of length $x_{i+1}-x_i=(y-x)/n$. By hypothesis, for every $i$, $$|f(x_i)-f(x_{i+1})|\leqslant M(x_{i+1}-x_i)^\alpha=M(y-x)^\alpha n^{-\alpha},$$ hence, by the triangular inequality, $$|f(x)-f(y)|\leqslant \sum\limits_{i=1}^n|f(x_i)-f(x_{i+1})|\leqslant M(y-x)^\alpha n^{1-\alpha}.$$ If $\alpha\gt1$, the RHS goes to zero when $n\to\infty$ because $n^{1-\alpha}\to0$, hence $f(x)=f(y)$, QED.

Hint: Show that $f'(y)$ exists and is equal to $0$ for all $y$. Then as usual by the Mean Value Theorem our function is constant.

HINT: Your idea is a good one. What happens when you divide the inequality by $|x-y|$?

It suffices to show that $f$ is differentiable and its derivative vanishes everywhere.

The hypothesis implies that, for every $x\in(a,b)$ and $h$, such that $x+h\in(a,b)$, we have that

$$ \frac{|f(x+h)-f(x)-0\cdot h|}{|h|} \le M\,|h|^{\alpha-1}. $$

The right hand side of the above tends to zero, as $h\to 0$, and therefore $f'(x)=0$!