Hölder exponents greater than 1 imply function to be constant?

This is not true for general metric spaces - if you have a function whose domain is a two-point space, then this function is $\alpha$-Hölder for every $\alpha>0$. $\alpha$-Hölder property for $\alpha>1$ implies the function being constant only in special spaces, like $\mathbb R^n$. Let me just focus on functions $f:\mathbb R\to\mathbb R$.

You mention "[the proofs] all make (at least indirect) use of some differentiability assumption on $f$". This is not quite correct - you don't need to assume differentiability, since $\alpha$-Hölder condition for $\alpha>1$ implies that the derivative exists - indeed, we just consider $$\left|\frac{f(x+h)-f(x)}{h}\right|\leq\frac{C|h|^\alpha}{|h|}\to 0,$$ so the limit exists and is zero everywhere. That way or another, we can prove that $f$ is constant directly as well: take any two points $x<y$ and let $x_0=x,x_1=x+\frac{y-x}{n},x_2=x+2\frac{y-x}{n},\dots,x_n=x+n\frac{y-x}{n}=y$. Then we have $$|f(x)-f(y)|\leq|f(x_0)-f(x_1)|+|f(x_1)-f(x_2)|+\dots+|f(x_{n-1})-f(x_n)|\\ \leq C|x_0-x_1|^\alpha+C|x_1-x_2|^\alpha+\dots+C|x_{n-1}-x_n|^\alpha\\ \leq n\cdot C\left|\frac{x-y}{n}\right|^\alpha=n^{1-\alpha}C|x-y|^\alpha\to 0,$$ so that $f(x)=f(y)$.