$S\subset\mathbb{R}^3$ compact, orientable, not a sphere $\Rightarrow K$ has positive and negative values 
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.
 A: 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).
A: I'm going to go ahead and take a crack at this. Since the surface $S$ is compact then any continuous function $f: S\to \mathbb{R}$ attains both a maximum and a minimum on $S$. Now recall that the Gauss curvature $K(p)$ is the determinant of the second fundamental form. If $K(p)\geq 0$ then there exists a point at which the second fundamental form is semi-definite positive.
In particular define the function $f:S\to \mathbb{R}$ by $f(p)=\lVert p-p_0\rVert^2$. Then $p$ is critical for the function $f$ if and only if $p_0\in p+ \lambda N(p)$ where $N(p)$ is the unit normal to the surface at $p$, $\lambda\in \mathbb{R}$. Next we compute the Hessian with $v$ a tangent vector: $(\frac{d^2f}{dt^2})_p(v)= \frac{d^2}{dt^2}\lvert_{t=0}\rVert\alpha(t) - p\lVert^2= \frac{d}{dt}|_{t=0}2\langle \alpha'(t), \alpha(t)-p_0\rangle$.
Differentiating one more time we thus obtain that this last expression equals $2\langle \alpha"(t), p-p_0\rangle+ \lVert v\rVert^2=2\langle (\lVert v\rVert^2 - \lambda \langle \alpha"(0), N(p)\rangle=2(\lVert v\rVert^2 -\lambda II_p(v))$. We thus have critical point for $f$ on $S$ because $S$ is compact. Let $p_1$ be the min and let $p_2$ be the max for $f$. Now as usual if we have a min the Hessian is positive definite if we have the max the Hessian is negative definite. So now grab the maximum since it works well: if you plug in $p_2$ you get that $\lambda II_p(v)\geq \lVert v \rVert^2$ (the scalar product if clearly positive definite, also notice we can choose $\lambda$ to be positive by picking the appropriate direction for the normal $N(p)$).
Thus there is indeed a point $p$ where the Gauss curvature is positive and thus the Gauss curvature is not always negative if $S$ is compact. Indeed you only need compactness.
