It is well known that if you perform a random walk on a 2 dimensional lattice then you will almost certainly reach every lattice point infinitely many times. Is the same result true if, instead of walking on a lattice, we walk in a random orientation (always using a distance of 1)?
Of course, we cannot expect to land on any given point with positive probability, so we modify the question to ask: in any disk, is the probability 1 that the random walk will eventually enter? I think this is equivalent to asking if the random walk will eventually any specific disk infinitely many times (e.g. one around the origin), because then there is a positive probability of taking any set of paths with a positive probability to get to the other disk, which is in theory not hard to construct.
What I have tried:
One approach is to use the result on the lattice (by converting a random walk on the plane to a random walk in the lattice) to prove this result, but I haven't made any meaningful progress in this direction.
Another approach is to mimic a proof that works over the lattice. The only proof strategy I am somewhat familiar with to prove the result over the lattice (although I am aware that there are others) is to show that the sum of the probabilities that you are at the origin after $n$ steps for each $n$ diverges, and then showing that this implies that the probability that you will return infinitely many times is 1. But it seems neither step generalizes directly. Something we could try is to prove that. for any point $P$ and any circle of radius $r$ containing $P$, the sum of the probabilities that the walk (re-)enters that circle on step $n$ is infinite; I think this would fix the second step to work in this case by using the argument from page 163 of https://services.math.duke.edu/~rtd/PTE/PTE4_1.pdf (the proof of theorem 4.2.2) by letting $r$ be half the radius of the original disk and always choosing a circle containing the origin.