 Solution

For $k\ge 2$, substitute $r_k=k^{-k}$ and $R=k$ into the exit formula: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 $\{S_{r_k}\le T_k\}$. Thus $\{{\tau }_0<\infty \}$ is contained in the liminf of these events. Fatou's inequality for their indicators gives probability at most ${\mathrm{liminf}}_{k}1/(k+1)=0$ for that liminf. Hence the polar point for planar Brownian motion conclusion isThe argument controls one event for all times, rather than merely proving zero hitting probability at each fixed time.