Andy Latto: I think this is really a theorem of differential geometry, not of algebraic geometry; the only place Warren's proof uses the fact that f is polynomial is when he needs it to be sufficiently smooth. --WDS: not so. I also used the fact the zero set of a polynomial, restricted to a compact region, has finite measure, e.g. cannot be an infinite spiral. One can easily construct an INFINITELY differentiable (i.e way smooth) function f(x,y) such that f(x,y)=0 is a spiral curve that within some annulus -- or disc -- makes an infinite number of winds. The function f will be 0 on the spiral, and in between windings will have a ridge or valley, but the ridge always has a slight slope, approaching but never quite becoming horizontal. As a result there are no critical points of f within the annulus.
Since we can replace R by any smaller simply connected region that doesn't touch C, we may as well choose R to be diffeomorphic to a disc, and the theorem then becomes:
For any sufficiently smooth (C2?) function f on the unit disc, if f is nonzero on the unit circle, but has zeroes inside the disc, then it has a critical point inside the disc.
Proof: if f is negative on the unit circle, its maximum is a critical point; if f is positive on the circle, its minimum is a critical point.
--a counterexample to that "theorem" and "proof" is the function f(x,y)=x? Oh, by "nonzero" you meant "never zero." Aha. Right, this is now trivial.
This works in any dimension, and we don't really need the Perelman-Thurston classification theorem, the Jordan Curve theorem, or the classification of surfaces.
Latto: All we need is the fact that the boundary of a compact simply connected set is connected; the theorem doesn't really require that R is simply connected, merely that it has connected boundary. WDS: my infinite spiral counterexample seems to indicate that is false (within a 3D solid torus)? Well, I think the final last winding of the spiral, which never comes, is a circular ring of critical points, which lie on the boundary of the region, but not inside it, and thus technically yield a counterexample, at least to some wordings of the theorem.