Let $(B_t)_{t \ge 0}$ be a $d$-dimensional Brownian motion. Then almost every $\omega$ in the underlying probability space $\Omega$ has this property: For each $\gamma > 1/2,$ the path $t \mapsto B_t(\omega)$ is nowhere Holder-continuous of order $\gamma$.
A function $f: I \to \mathbb{R}^d$ defined on an interval $I \subset \mathbb{R}$ is called Holder-continuous of order $\gamma>0$ at a point $t_0 \in I$, if there exists a finite constant $c$, possibly depending on $t_0$, such that $$|f(t)-f(t_0)| \le c|t-t_0|^\gamma$$ for all $t \in I$ lying in a sufficiently small neighborhood of $t_0$. If for a fixed $\gamma>0$ the above condition of Holder continuity is satisfied at no point $t_0 \in I$, then $f$ is said to be nowhere Holder continuous of order $\gamma$.
I am looking at the proof of this theorem from Bauer's Probability Theory.
It says here that it suffices to prove the analogous statement for a fixed (but arbitrary) $\gamma > 1/2.$ For if that has been done then to each such $\gamma$ will be associated an event $\Omega_\gamma \in \mathscr{A}$ with $P(\Omega_\gamma)= 0$ such that for $\omega \in \Omega \setminus \Omega_\gamma$ the path $t \mapsto B_t(\omega)$ is nowhere Holder-continuous of order $\gamma$. The union of the $\Omega_\gamma$ over all $\gamma \in (1/2,+\infty) \cap \mathbb{Q}$ is an event $\Omega_0 \in \mathscr{A}$ with $P(\Omega_0)=0$, and every path $t \mapsto B_t(\omega)$ with $\omega \in \Omega \setminus \Omega_0$ has the claimed property.
Why, is the path nowhere Holder-continuous of any order $\gamma > 1/2$, if it is nowhere Holder-continuous for any rational $\gamma > 1/2$?
If a function is Hölder-continuous of order $\gamma$ at a point, then it is also Hölder-continuous of order $\gamma'$ for any $\gamma'<\gamma$. This is immediate because $|t-t_0|^{\gamma'}\geq |t-t_0|^\gamma$ as long as $|t-t_0|\leq 1$, and for Hölder continuity at $t_0$ you only care about what happens when $|t-t_0|$ is sufficiently small.
So, since there are rational numbers arbitrarily close to but greater than $1/2$, failure of Hölder continuity for all rational orders greater than $1/2$ implies failure of Hölder continuity for all orders greater than $1/2$.