= Solution
Translate the centre of the ball to zero. Choose an inner radius $\rho$ strictly smaller than the ball's radius. If the starting point already lies in the ball there is nothing to prove; otherwise write $d=|z|>\rho$ and choose $R>d$. Let $H_\rho,H_R$ be the first hits of the two circles.
The <Brownian exit time> from the annulus $\rho<|z|<R$ is finite almost surely: it is no later than exit of the first coordinate from $(-R,R)$, which is finite by part (a). The function $\log|z|$ is <harmonic> on this annulus. The <Itô formula> and <optional stopping theorem> apply to its bounded stopped values, yielding
$$
\log d
=\mathbb P_z(H_\rho<H_R)\log\rho
+\mathbb P_z(H_R<H_\rho)\log R.
$$
Solving gives the <planar Brownian annulus hitting probability>
$$
\boxed{\mathbb P_z(H_\rho<H_R)=\frac{\log R-\log d}{\log R-\log\rho}.}
$$
As $R\to\infty$ this tends to one, while the event in the formula is contained in $\{H_\rho<\infty\}$. Thus the inner circle is hit almost surely. That circle lies strictly inside the original open ball, proving the <recurrence of planar Brownian motion> in the form requested:
$$
\boxed{\mathbb P_z(\text{hit the given nonempty open ball})=1.}
$$
A probability-one intersection over balls with rational centres and positive rational radii makes this conclusion simultaneous for all nonempty open balls, since every such ball contains one from that countable family.
\b[Using a smaller inner circle ensures an actual visit to the open ball, rather than only a hit of its boundary.]
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