The proof defines F(X) as the set of compact subsets of X (here the unit square). The compactness of F(X) follows from (necessary?) the compactness of its elements, so we need to find a topology where all our allowable symbols are compact, and this is not the standard topology if you allow the spiral. If you take another topology (equivalently space of symbol descriptions) then you again must show that this compactness still corresponds to the desired notion of "distinguishability".
My topology is rusty and I don't genuinely doubt the validity of the argument, but I'm having fun trying to poke holes.
Your example centered on a symbol which, if viewed as a subset of the plane, is not compact. I tried to argue that the set of symbols that you describe (ink of varying levels of intensity in the unit square) still is a compact set, even though the symbols themselves are no longer represented by compact sets.
My topology is rusty and I don't genuinely doubt the validity of the argument, but I'm having fun trying to poke holes.