Perhaps this is what you're looking for?

For each $t>0$, $B_t/\sqrt{t}$ has a standard normal distribution. Therefore for any $k$ we have $\mathbb{P}(B_t/\sqrt{t} \ge k) = 1-\Phi(k) > 0$ where $\Phi$ is the standard normal cdf. So if we fix a sequence $t_n \downarrow 0$ and let $X = \limsup_{t_n \downarrow 0} B_{t_n}/\sqrt{t_n}$ we have $\mathbb{P}(X \ge k) \ge 1-\Phi(k) > 0$. But by the Blumenthal 0-1 law, $X$ is almost surely constant, so $X \ge k$ almost surely. Now $k$ was arbitrary so $X = +\infty$ almost surely. In particular, almost surely, $\sup_{t > 0} B_t/\sqrt{t} = +\infty$.

Perhaps this is what you're looking for?

Fix a sequence $t_n \downarrow 0$ and some $k > 0$ and let $A_n = \{ \sup_{0 < t < t_n} B_t/\sqrt{t} \ge k\}$. Since $B_{t_n}/\sqrt{t_n}$ has a standard normal distribution, we have $$\mathbb{P}(A_k) \ge \mathbb{P}(B_{t_n}/\sqrt{t_n} \ge k) = 1-\Phi(k) > 0$$ where $\Phi$ is the normal cdf.

Now $A_1 \supseteq A_2 \supseteq \cdots$ so if $A = \bigcap_n A_n$, then by "continuity from above" we have $$\mathbb{P}(A) = \lim_{n \to \infty} \mathbb{P}(A_n) \ge 1-\Phi(k) > 0.$$ But $A$ is in the "germ field" $\mathcal{F}_0^+$ so the Blumenthal 0-1 law says we must have $\mathbb{P}(A) = 1$. On the event $A$ we have $\sup_{0 < t < 1} B_t/\sqrt{t} \ge k$, and $k$ was arbitrary, so almost surely we have $\sup_{0 < t < 1} B_t/\sqrt{t} = +\infty$.