Can a plane be split into three connected sets so that $\epsilon$-neighbourhood of any point of any one set also contains points of two other sets?

Question

Math SE.

This question was a shower thought of mine. I tried to come up with an answer by twisting comb spaces and cantor sets, but to no avail. I was educated as experimental physicist, so I probably don't know enough math to come up with a working solution. But this problem is like a splinter in my mind. It drives me crazy for 11 hours straight. Please, help!

EDIT: I should have added that the condition about $\epsilon$-neighbourhood must be symmetric for A, B, C. That is,

1. $\epsilon$-neighbourhood of any point in A must contain at least one point of B and at least one point of C;
2. $\epsilon$-neighbourhood of any point in B must contain at least one point of A and at least one point of C;
3. $\epsilon$-neighbourhood of any point in C must contain at least one point of A and at least one point of B.

I have also modified the title to reflect this edit. Before this edit the title was: "Can a plane be split into three connected sets A, B and C so that $\epsilon$-neighbourhood of any point of A also contains points of B and C?"

Answer 1

You can even have continuum many such sets. By transfinite induction, construct a partition $\{X_i : i < 2^{\omega}\}$ of plane such that each $X_i$ meets every perfect subset of plane. This ensures connectedness because if $X$ meets every perfect set in plane and if $U, V$ are non empty disjoint open sets then $\mathbb{R}^2 \backslash (U \cup V)$ cannot be countable (since cocountable subsets of plane are (even path) connected) so $X$ meets it.