Assume f is a continuous function on the (unit) square in real plane. Name the edges N,S,E and W in the natural way. Assume f is >0 at W edge and <0 on E edge.
Intuitively it is clear that there is (at least one) level curve for level=0 that connects N edge to S edge.
Without requiring more smoothness than just continuity, I think I can prove that the open set where abs(f) < eps (small eps>0) always contains a connecting path, with a finite set of horizontal and vertical legs.
This seems like a very natural generalization of 1-D intermediate value theorem. Are there elementary "well-known" textbook results (meaning with an elementary short proof fitting to calculus or analysis, not advanced topology)? Is there any "obvious" proof only requiring continuity, no differentiability?