= Solution
First fix a deterministic point $y$ different from the initial point. By translation, work with $y=0$ and $B_0=x\ne0$. For $R>|x|$, hitting $0$ before $T_R$ requires hitting every circle of radius $r\in(0,|x|)$ before $T_R$. Part (a) gives
$$
\mathbb P_x(T_0<T_R)\leq\frac{\log R-\log|x|}{\log R-\log r}\longrightarrow0\qquad(r\downarrow0).
$$
Any finite-time hit of $0$ lies before exit from some sufficiently large integer-radius circle, since a continuous path on a finite interval is bounded. A countable union over such radii therefore shows that this fixed point is never hit. This is the <polar point for planar Brownian motion> property; it is a statement about each fixed point, not about avoiding all points simultaneously.
For the initial point $x$, fix a deterministic $\varepsilon>0$. The <Gaussian distribution> of $B_\varepsilon$ has a density, so $B_\varepsilon\ne x$ almost surely. Conditional on $\mathcal F_\varepsilon$, the future is <Brownian motion> starting from $B_\varepsilon$, and the preceding fixed-point result shows that it never hits $x$. Taking $\varepsilon=1/m$ and a countable union proves \b[there is no return to the starting point at any positive time].
In contrast, neighborhoods are revisited arbitrarily late. Choose $r>0$ so that the closed disk centered at $x$ of radius $r$ lies in the given open set $U$. From any point $z$ outside this disk, let $R\to\infty$ in part (a), centered at $x$. It gives
$$
\mathbb P_z(T_r<\infty)\geq\lim_{R\to\infty}\frac{\log R-\log|z-x|}{\log R-\log r}=1.
$$
At each deterministic integer time $n$, condition on $B_n$. If it is in the disk there is already a visit to $U$ at time $n$; if it is outside, the <Markov property> and the last calculation ensure a later visit to that disk, hence to $U$. With <probability> one this holds for all $n$ simultaneously. Thus \b[the set of visits to $U$ is unbounded]. The <recurrence of planar Brownian motion> concerns neighborhoods even though every fixed point is polar.
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