= Solution
\b[True.] The preceding distributional argument alone shows that almost-sure convergence to zero cannot hold with <probability> one. To establish almost-sure nonconvergence, rather than merely failure of a probability-one convergence claim, use the <Kolmogorov zero-one law>.
Let $A=\{B_n/\sqrt n\to0\}$. For every fixed $m$, subtracting $B_m/\sqrt n\to0$ shows that
$$
A=\left\{\frac{\sum_{k=m+1}^n(B_k-B_{k-1})}{\sqrt n}\longrightarrow0\right\}.
$$
Thus $A$ belongs to the <tail sigma-algebra> of the independent unit-time increments; it is a <tail event>. The <Kolmogorov zero-one law> gives $\mathbb P(A)\in\{0,1\}$. If its <probability> were one, then $B_n/\sqrt n\to0$ in <probability>, contrary to its standard <normal distribution> at every $n$. Therefore $\mathbb P(A)=0$. Continuous-time convergence would imply integer-time convergence, so
$$
\boxed{\mathbb P\bigl(B_t/\sqrt t\text{ does not converge to }0\bigr)=1.}
$$
The more general <Brownian fluctuations exceed the square-root scale> result is consistent with this direct tail-event proof. No <independence> of the normalized values themselves has been assumed.
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