Here's an almost complete proof, using ingredients from Tom's proof (but discarding #4). First argue that there must be a way to turn the initial "nested squares" multipath configuration into a maximal single-path configuration via successive demotions (see my previous email). Then, claim that at least one cell in each of the k-1 outermost rings must be demoted in the process. For, if not, then we'd get a non-simply-connected band of cells of constant winding-number at the end of the process, which would disconnect the multipath into two parts (one "outer" and one "inner"). The only gap I see here is, why couldn't the outer part or the inner part of the maximal single-path configuration be empty? The answer must be, This woud contradict maximality! I think I see a way to fill the gap (two cases: one for the outer part and one for the inner part). The first I'm 90% sure of; the second I'm less sure of. My approach is kind of fiddly. Can anyone find a more conceptual way of wrapping up the proof? Jim