$f$ Holder continuous with Holder exponent $p>1\implies f \text{ is constant}$

Your approach of splitting the interval is essentially correct. Here it is in detail.

Let $x$ and $y$ be any two distinct points in $I$. Without loss of generality let us suppose $x <y$. Splitting $[x, y]$ in $n$ intervals of length $\frac{|y-x|}{n}$, we have $n+1$ points $x_k$, such that $x=x_0 < x_1 < \dots <x_{n-1} < x_n=y$ and $|x_k -x_{k-1}| = \frac{|y-x|}{n}$. So we have

\begin{align*} d_{Y}(f(y), f(x))&\leq\sum_{k=1}^{k=n}d_{Y}(f(x_k), f(x_{k-1})) \leq C\cdot\sum_{k=1}^{n} |x_k-x_{k-1}|^p = \\ &=C\sum_{k=1}^{n} \left(\frac{|y-x|}{n} \right)^p =C|y-x|^p n^{1-p} \end{align*}

Now, note that as $n \to \infty$, we have $C|y-x|^p n^{1-p} \to 0$, and so you have that $d_{Y}(f(y)- f(x))=0$. So $f(y)=f(x)$. Since, $y$ and $x$ are arbitrary points in $I$, it follows that $f$ is constant.