= Solution
Let $B$ be a standard <planar Brownian motion> started at $z$. Neighbourhood recurrence means that, with <probability> one, every nonempty open subset of $\mathbb R^2$ is visited at arbitrarily large times. Not hitting points means that for each fixed $w\ne z$, $\mathbb P_z(\exists t\geq0:B_t=w)=0$. If $w=z$, time zero must be excluded, and there is still no return at any positive time. This assertion concerns each prescribed point; it does not say that the path contains no points.
We prove both claims from the <planar Brownian annulus hitting probability>. Fix a center $w$ and radii $0<r<\rho=|z-w|<R$, and define the <stopping times> $\tau_r=\inf\{t:|B_t-w|=r\}$ and $\tau_R=\inf\{t:|B_t-w|=R\}$. Their minimum $\sigma$ is finite <almost surely>: a coordinate of <Brownian motion> cannot stay forever in a bounded interval, by the <Brownian reflection principle>, so exit from the outer disk is finite. On the annulus,
$$
\Delta\log|x-w|=0.
$$
The <Itô formula> therefore makes $\log|B_{t\wedge\sigma}-w|$ a <local martingale>. It is bounded between $\log r$ and $\log R$, hence is a genuine <martingale>. The <optional stopping theorem> at $t\wedge\sigma$, followed by the <bounded convergence theorem>, gives
$$
\log\rho=\mathbb P_z(\tau_r<\tau_R)\log r+\mathbb P_z(\tau_R<\tau_r)\log R.
$$
Thus
$$
\boxed{\mathbb P_z(\tau_r<\tau_R)=\frac{\log(R/\rho)}{\log(R/r)}.}
$$
The exit can only be at one of the two distinct boundaries by path continuity.
For a fixed $r$, let $R\to\infty$. These events increase to $\{\tau_r<\infty\}$: any path up to a finite hitting time is bounded and so avoids some sufficiently large outer circle. The ratio tends to one, proving that any closed disk of positive radius is hit <almost surely> from any starting point outside it. An initial point inside it already counts as a hit.
To prove the stronger recurrence statement, take a closed disk contained in any prescribed nonempty open set. At each integer time $n$, the <Markov property> and the preceding disk-hitting result imply that some time $t\geq n$ visits this disk with <probability> one. Intersecting these events over all $n$ proves visits at arbitrarily large times. A <countable> basis of disks with rational centers and rational radii makes this simultaneous for all nonempty open sets. This is <recurrence of planar Brownian motion>, or equivalently neighbourhood recurrence.
For point avoidance, keep $R>\rho$ fixed and let $r\downarrow0$. Hitting $w$ before $\tau_R$ requires hitting every inner circle first. The displayed ratio tends to zero, so this point-hitting event has <probability> zero. Any finite hit of $w$ would occur before exit from some outer circle with integer radius, again because the preceding path is bounded. A <countable union> now gives $\boxed{\mathbb P_z(T_w<\infty)=0\ (z\ne w)}$, the <polar point for planar Brownian motion> assertion.
Finally, even if the process starts at $w$, for each deterministic $s>0$ its <normal distribution> has a density, so $\mathbb P(B_s=w)=0$. The <Markov property> at $s$ and the result from other starting points exclude any visit to $w$ at or after $s$. Taking $s=1/n$ excludes all positive-time returns. Thus \b[planar Brownian motion repeatedly approaches every point, while almost surely never hitting any prescribed point at positive time].
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