= Solution
For $k\geq2$, substitute $r_k=k^{-k}$ and $R=k$ into the exit formula:
$$
\boxed{\mathbb P(S_{r_k}\leq T_k)=\frac{\log k}{\log k+k\log k}=\frac1{k+1}\longrightarrow0.}
$$
Let $\tau_0$ be the first hit of the origin. On any path with $\tau_0<\infty$, continuity makes the radius bounded on $[0,\tau_0]$. For all sufficiently large $k$, that path reaches radius $r_k$ before $\tau_0$ and has not yet reached radius $k$, so it belongs to $\{S_{r_k}\leq T_k\}$. Thus $\{\tau_0<\infty\}$ is contained in the liminf of these events. Fatou's inequality for their indicators gives probability at most $\liminf_k1/(k+1)=0$ for that liminf. Hence the <polar point for planar Brownian motion> conclusion is
$$
\boxed{\mathbb P(|B_t|>0\text{ for every }t\geq0)=1.}
$$
The argument controls one event for all times, rather than merely proving zero hitting probability at each fixed time.
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