= Solution
First suppose $|x+z|\leq r$ for some lattice point $-z$. If $|x+z|=r$, the desired hit is already at time zero. If $|x+z|<r$, finite exit from that bounded <open ball> and continuity give the required <sphere> hit <almost surely>.
It remains to consider $|x+z|>r$ for every $z\in\mathbb Z^3$. At each integer time $n$, choose a nearest lattice point $-z_n$ to $B_n$ by rounding its three coordinates, with a fixed rule at ties. Then $|B_n+z_n|\leq\sqrt3/2$. Put $\delta=\min(r/2,1/4)>0$. Conditional on the past, $B_{n+1}-B_n$ is a standard three-dimensional <Gaussian vector>. Its <probability density function> is bounded below on the <open ball> of radius $\delta$ centered at $-(B_n+z_n)$ by
$$
(2\pi)^{-3/2}\exp\!\left[-\tfrac12(\sqrt3/2+\delta)^2\right].
$$
Multiplying by the volume of that <open ball> gives a constant $\varepsilon>0$, independent of $n$ and the past, such that
$$
\mathbb P\bigl(|B_{n+1}+z_n|<\delta\mid\mathcal F_n\bigr)\geq\varepsilon.
$$
The probability of avoiding all lattice-centered radius-$r$ <open balls> at the first $N$ positive integer times is therefore at most $(1-\varepsilon)^N$, by iterated <conditional expectation>. Eventually the path enters one such <open ball> <almost surely>. Since the initial point was outside that particular <open ball>, continuity forces an earlier crossing of its <sphere>. This proves <Brownian hitting of lattice spheres>:
$$
\boxed{\mathbb P_x\!\left(\exists t\geq0,\ z\in\mathbb Z^3:\ |B_t+z|=r\right)=1.}
$$
A uniform chance of hitting some member of the periodic family replaces recurrence to any one prescribed <sphere>; there is no contradiction with (b).
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