In the proof of theorem 8.1.3 of Durrett's book Probability:Theory and Examples, there is a bit which I don't understand.
I don't get how we arrived at the last inequality with just triangle inequality. I can see $|X(q)-X(r)|\le \sum_{i=1}^n|X(s_i)-X_(s_{i-1})| \le n \delta(\omega)^\gamma $ but that's about it .
Choose $n \in \mathbb{Z}$ in such a way that $$\frac{\delta(\omega)}{2} \leq \frac{1}{2^n} \leq \delta(\omega) \tag{1}$$ and set $s_i := q + (r-q) i 2^{-n}$, $i=0,\ldots,2^n$. Then
$$|X(s_i)-X(s_{i-1})| \leq A |s_i-s_{i-1}|^{\gamma} = A (2^{-n})^{\gamma} |r-q|^{\gamma}$$
which implies, by the triangle inequality,
$$|X(q)-X(r)| \leq \sum_{i=1}^{2^n} |X(s_i)-X(s_{i-1})| \leq A 2^n (2^{-n})^{\gamma} |r-q|^{\gamma} = (2^n)^{1-\gamma} |r-q|^{\gamma}.$$
If $1-\gamma \leq 0$, this proves the assertion. If $1-\gamma>0$, then we note that, by $(1)$, $2^n \leq 2/\delta(\omega)$, and therefore
$$|X(q)-X(r)| \leq \left( \frac{2}{\delta(\omega)} \right)^{1-\gamma} |r-q|^{\gamma} = C(\omega) |r-q|^{\gamma}.$$