if you are not familiar with the concept, s-Lipschitz continuous functions are functions such that
for all x, y $\in$ the domain of $f$, $\exists $ a constant C s.t. $|f(x)-f(y)|<C|x-y|^s$ for all s$\in$[0,1]
I am not sure how to prove that s cannot be larger than 1, otherwise, the function would have to be constant functions. And whether the domain of the function needs to be a compact set, a bounded set, or any set is ok?
I barely know that to move |x-y| to the left side of the inequality, it would be
$f'(y)<C|x-y|^{s-1}$ when x is close enough to y.
As @Arctic Char has pointed out, usually the definition requires that $x,y$ are quantified after $C$ (so that the constant $C$ is independent of $x$ and $y$).
You must be careful, the derivative is the limit $f'(y)=\lim_{x \to y} \frac{f(x)-f(y)}{x-y}$; so, after you divided and taken the limit as $x \to y$, you cannot take the limit only on the left hand side leaving the right hand side $|x-y|^{s-1}$ unchanged (it depends on $x$ too).
From: $$\left|\frac{f(x)-f(y)}{x-y}\right| \le C|x-y|^{s-1}$$ If $s >1$, it is $s-1>0$ and so, taking the limit as $x \to y$, it is $f'(y)=0$ for each $y$. Hence, if $x,y$ belong to an open connected set (for instance, an interval of $\mathbb{R}$) , you can deduce that $f$ is constant because its derivative is $0$ on an open connected set.