Suppose that for every $K > 0$ we have that $$ \mathrm{P} \left [\inf \left\{t >0: \left| B_t \right|> K \sqrt{t} \right \} = 0 \right] = 1, $$ where $B$ is a Brownian Motion.
I would like to motivate - from an explicit definition of Hölder-$\tfrac{1}{2}$ continuity - that this means that almost surely $B$ is not Hölder-$\tfrac{1}{2}$ continuous at $t$.
Perhaps it is possible to show that $B$ isn't Hölder-$\tfrac{1}{2}$ continuous "from the right". This would be sufficient for the claim to be true and perhaps easier than my attempt at a solution below, since it would be easier to apply the property $B_t - B_s \sim B_{t-s}$?
Here comes my long winded attempt to a solution:
Given any $t$ we have that the path $B_s(\omega)$ is $\tfrac{1}{2}$-Hölder continuous at $t$ if there exists a $K>0$ and $\epsilon>0$ such that for $$ |t-s| < \epsilon, $$ it is the case that $$ \left | B_t(\omega) - B_s(\omega) \right | \le K \sqrt{|t - s|}. $$ We will try to use the fact that $$ B_t - B_s \sim B_{t-s}. $$ If we want to calculate the probability that $B$ is $\tfrac{1}{2}$-Hölder continuous at $t$, we have to consider that the $\epsilon$ in the above requirement may differ for different paths (the $K$ may of course also vary).
My idea was to start with an explicit expression for the event that $B_t$ is Hölder-$\tfrac{1}{2}$ continuous at $t$, and then show that the given probability means that the complement of this event has probability one. Thus the event that $B$ is $\tfrac{1}{2}$-Hölder continuous at $t$ may be written as, $$ \bigcup_{K=1}^\infty \bigcup_{n=1}^\infty\left \{\left | B_t - B_s \right | \le K \sqrt{|t - s|}:0< |t-s|<\frac{1}{n} \right \}. $$ Now, $$ \left \{\left | B_t - B_s \right | \le K \sqrt{|t - s|}:0< |t-s|<\frac{1}{n} \right \} $$ is distributed as (Not $100 \%$ sure about this) $$ \left \{\left | B_{|t-s|}\right | \big/ \sqrt{|t - s|} \le K :0< |t-s|<\frac{1}{n} \right \}, $$ and thus we may consider the event where $B$ is $\tfrac{1}{2}$-Hölder continuous at $t$ as \begin{multline*} \bigcup_{K=1}^\infty \bigcup_{n=1}^\infty\left \{\left | B_{|t-s|}\right | \big/ \sqrt{|t - s|} \le K :0< |t-s|<\frac{1}{n} \right \} \\ = \bigcup_{K=1}^\infty \bigcup_{n=1}^\infty \left \{ \sup_{0 < |t-s| < \frac{1}{n} }\left |B_{|t-s|}\right | \big/ \sqrt{|t-s|} \le K \right \} \\ =\bigcup_{K=1}^\infty \bigcup_{n=1}^\infty \left \{ \sup_{0 < t < \frac{1}{n} }\left |B_{t}\right | \big/ \sqrt{t} \le K \right \}. \end{multline*}
If the above is correct, one may consider (a set that is distributed as) a subset of the the event that $B$ isn't $\tfrac{1}{2}$-Hölder continuous at $t$. We have that $$ \left (\bigcup_{n=1}^\infty \left \{ \sup_{0<t \le \frac{1}{n} }\left |B_{t}\right | \big/ \sqrt{t} \le K \right \}\right ) ^C = \left \{\inf \left\{t >0: \left| B_t \right|> K \sqrt{t} \right \} = 0 \right \}. $$ Now, $$ \left \{\inf \left\{t >0: B_t > K \sqrt{t} \right \} = 0 \right \} \subset \left \{\inf \left\{t >0: \left| B_t \right|> K \sqrt{t} \right \} = 0 \right \}, $$ and so the given condition is sufficient for the desired conclusion as we have that a subset of the set where $B$ isn't Hölder-$\tfrac{1}{2}$ continuous has probability one.
Most grateful for any help provided!
Not clear from the question if you are having trouble proving the condition: \begin{align} P (\inf \{t>0\mid |B_t| > K\sqrt{t}\}=0)= 1 \text{ for all }K>0. \label{eq1} \tag{1} \end{align}
But if you assume \eqref{eq1}, showing that $B$ is not Holder-$\frac{1}{2}$ continuous is not difficult. Clearly by the definition of Holder continuity, this condition says that $B$ is not Holder-$\frac{1}{2}$ continuous at $t=0$.
And now apply the Markov property of $B$. Fix $t>0$. If $B$ is a standard Brownian motion then so is the process $W_s = B_{s+t}-B_t$. Thus condition \eqref{eq1} holds for $(W_s)_{s\ge 0}$ and it is not Holder-$\frac{1}{2}$ continuous at $s=0$. This implies that $B$ is not Holder-$\frac{1}{2}$ continuous at $t$.