Let $S$ be a compact, orientable surface embedded in $\mathbb{R}^3$, not diffeomorhic to $\mathbb{S}^2$. Prove that its Gauss curvature attains positive and negative values (consequently, it eventually vanishes).
Since $S$ is not a sphere, the only possible values for $\chi(S)$ are $0,-2,-4,-6,...$, so by the Gauss Bonnet theorem, $\int\int_S Kd\sigma=2\pi\chi(S)\leq 0$, which means $K$ cannot be always positive. I'm trying to prove that $K$ also cannot be always negative, but I'm having trouble with it.
Intuitively, the fact that $S$ is compact should be enough to guarantee that, but I have no idea how to implement the idea.
Enclose the surface in a sphere of sufficiently large radius, then shrink the sphere contiuously until it touches the surface for the first time. The point(s) where that happens have positive Gauss curvature (as least the curvature of the sphere).