$f$ is a real function and it is $\alpha$-Holder continuous with $\alpha>1$. Is $f$ constant? Maybe this is a well know result, however, I could not find it. Before stating it, let me write here a well know result (at least for me)

Assume that $\Omega\subset\mathbb{R}^N$ is a open domain and $f:\Omega\to\mathbb{R}$. If there is constants $L>0$ and $\alpha>1$ such that $$|f(x)-f(y)|\leq L |x-y|^\alpha,\ \forall\ x,y\in\Omega$$
  then, $f$ is constant in each connected componente of $\Omega$.

The above result can be proved, for example, by showing that $\nabla f=0$ and then we join points in the same connected component by a continuous curve.
Now my question is: 

Assume that $\Omega\subset\mathbb{R}^N$ and $f:\Omega\to\mathbb{R}$. Suppose that there is constants $L>0$ and $\alpha>1$ such that $$|f(x)-f(y)|\leq L |x-y|^\alpha,\ \forall\ x,y\in\Omega$$
  Can we conclude that $f$ is constant in each connected componente of $\Omega$?

Maybe it is necessary to add the hypothesis that each connected component of $\Omega$ is pathwise connected? 
Remark: Note that in the question, $\Omega$ does not need to be a open set. It is now any set.

Claim: For $f$ as stated, and $p > 1$, assume that the region $\Omega$ is path connected by continuous paths of finite total variation. Then $f$ is constant on $\Omega$.
To see this, let $x$, $y$ be given, and choose a path $v(t) : [0,1]\rightarrow \Omega$ of finite arc length $l(v)$ that connects $x$ to $y$. Without loss of generality assume $l(v)\ne 0$ (otherwise $f(x)=f(y)$.) Let $\epsilon > 0$ be given. Find a partition
$$
               {0=t_{0} < t_{1} < \cdots t_{n}=1}
$$
refined enough so that $|v(t_{j-1})-v(t_{j})| < (\epsilon/2l(v))^{1/(p-1)}$ for all $j$, which is possible because $x$ is continuous on $[0,1]$ and, hence, also uniformly continuous on $[0,1]$. Then
$$
\begin{align}
    |f(x)-f(y)| & \le \sum_{j=1}^{n}|f(v(t_{j-1}))-f(v(t_{j}))| \\
                & \le L\sum_{j=1}^{n}|v(t_{j-1})-v(t_{j})|^{p-1+1} \\
                & \le \frac{\epsilon}{2l(v)}\sum_{j=1}^{n}|v(t_{j})-v(t_{j-1})|\le \frac{\epsilon}{2} < \epsilon.
\end{align}
$$
Because $\epsilon$ was arbitrary, then $f(x)=f(y)$. This is true for all points $y$ connected to $x$ by such paths, which is everything in this case. So $f$ is constant on $\Omega$.
