= Solution
Since $K\geq0$, <Gauss-Bonnet theorem>[Gauss-Bonnet] gives $4\pi(1-g)\geq0$, hence $g\leq1$. If $g=1$, the integral of the continuous nonnegative function $K$ is zero, so $K$ vanishes everywhere.
That is impossible for a compact regular surface in $\mathbb R^3$. Choose $p\in S$ maximizing the <Euclidean distance> from the origin, after translating the origin off the surface if necessary. The sphere centered at the origin through $p$ is a <supporting sphere>. The <supporting-sphere curvature bound> shows that both <principal curvatures> at $p$ have the same sign and nonzero magnitude, so $K(p)>0$. Therefore $g\ne1$, and $g=0$. The <classification theorem for surfaces> now shows that $S$ is diffeomorphic to the sphere.
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