$f$ is a real function and it is $\alpha$-Holder continuous with $\alpha>1$. Is $f$ constant?

Counterexample

As usual, studiosus is right: the answer is negative. A natural parametrization of an arc of the von Koch snowflake gives a topological embedding $g:[0,1]\to \mathbb R^2$ such that $$|g(x)-g(y)|\ge C|x-y|^{p},\quad p=\frac{\log 3}{\log 4}$$ for all $x,y\in [0,1]$, with $C$ independent of $x,y$. The inverse $f=g^{-1}$ is a continuous map from a curve to $[0,1]$ which is Hölder continuous with exponent $1/p>1$.

More generally, for every $p\in (0,1)$ there is a topological embedding $g$ of $\mathbb R $ into a Euclidean space such that $$C |x-y|^p\le |g(x)-g(y)|\le C'|x-y|^{p}$$ for all $x,y\in\mathbb R $. This can be constructed directly, or obtained as a special case of Assouad's embedding theorem. The inverse of $g$ is Hölder continuous with exponent $1/p$ which can be arbitrarily large.

Positive result

To conclude that $f$ is constant, you need an additional geometric (not just topological) assumption on $\Omega$. It suffices to assume that $\Omega$ is quasiconvex (there is $C$ such that every two points $x,y\in \Omega$ can be joined by a curve of length at most $C|x-y|$).

Here is a weaker assumption: for every $x,y\in \Omega$ there is a connected set $E\subset \Omega$ that contains both $x$ and $y$ and has Hausdorff dimension less than $\alpha$.

Proof. Given $x,y\in\Omega$, take $E$ as above. The behavior of Hausdorff dimension under $\alpha$-Hölder maps is well known: $\operatorname{dim} f(E)\le \alpha \operatorname{dim} E$. Hence $\operatorname{dim} f(E)<1$. On the other hand, $f(E)$ is a connected subset of $\mathbb R$, i.e., either a point or an interval. Thus, $f(E)$ is a point, and $f(x)=f(y)$. $\quad\Box$