The Kolmogorov continuity theorem can be stated as
If $(X_t)_{t \geq 0}$ is a stochastic process valued in a complete metric space for which there exists constants $q,\epsilon,C > 0$ such that $$ \forall s,t \geq 0,~~E[d(X_s,X_t)^q] \leq C|t-s|^{1+\epsilon}, $$ then there exists a modificiation of $X$ whose paths are locally $\gamma$-Hölder continuous for any $\gamma \in (0,\epsilon/q)$.
I followed the classical proof in Jean-François Le Gall's "Brownian Motion, Martingales, and Stochastic Calculus" where it is proved that
If $(X_t)_{t \in [0,1]}$ is a stochastic process valued in a complete metric space for which there exists constants $q,\epsilon,C > 0$ such that $$ \forall ~t,s \in [0,1],~~E[d(X_s,X_t)^q] \leq C|t-s|^{1+\epsilon}, $$ then there forall $\gamma \in (0,\epsilon/q)$ there exists a modificiation of $X$ whose paths are $\gamma$-Hölder continuous.
For a fixed $\gamma$ I see that $[0,1]$ doesn't play any role and can be replaced by any bounded intervall. Then I see that we can get the unbounded version taking a partition of $\mathbb R^+$ by bounded intervalls and conciliating the obtained modifications.
I dont see why we can take the modification to locally have any Hölder constant in $(0,\epsilon/q)$. Perhaps denote $\tilde X^\gamma$ the extension who locally has $\gamma$ Hölder constant and take the limit $\gamma \rightarrow \epsilon/q$ but I don't see how such a limit would be taken.
The key fact is that we can always not distinguish a countable number of continuous stochastic processes. Then taking $\gamma_n \in (0,\epsilon/q)$ increasing to $\epsilon/q$ and not distinguishing the corresponding processes we are done.