Let $\alpha > 1$ and $c \in \mathbb{R}_{+}$. If $f: U \rightarrow \mathbb{R}^{n}$, where $U \subseteq \mathbb{R}^{m}$ is an open set, satisfies $|f(x) - f(y)| \leq c|x-y|^{\alpha}$ for any $x,y \in U$, then $f$ is constant on each connected component of $U$.
The theorem above is a slight generalization of the result seen here, since we have a function from $\mathbb{R}^{m}$ to $\mathbb{R}^{n}$. I've seen some proofs of this result and I noticed that many people try to avoid differentiability. On the link I provided above, the accepted answer uses derivatives and is supposedly wrong (?). It's got me second-guessing my own approach since it's pretty close to what I had in mind. So, I want to figure out where I might have made a mistake and what the common pitfalls are when using differentiability in these types of proofs.
Attempt: Let $x \in U$, $v \in \mathbb{R}^{m}$ and $t$ sufficiently small such that $x + tv \in U$. Then, $$0 \leq \frac{|f(x+tv) - f(x)|}{|t|} \leq c |t|^{\alpha - 1} |v|^{\alpha}.$$ As $t$ goes to zero, the limit above exists and is equal to zero. Therefore, $\partial_v f(x) = 0,\forall v \in \mathbb{R}^{m}$ and $\forall x \in U$. Note that, in particular, each component of $f$, namely $f_j, j = 1, \ldots, n$, has vanishing directional derivatives on every point. Thus, we can apply the MVT component-wise on any connected component $C$ of $U$ to conclude that $f$ must be constant on $C$.
