So if we have a compact surface $S \subset \mathbb{R}^3 $ then the function $f:S\rightarrow \mathbb{R}_{\geq 0 },\ x \mapsto ||x|| $ has a maximum value say $R >0$ at $p \in S$. Then we must have the surface $S$ fully enclosed by the sphere of radius $R$ as otherwise $f(p)$ is not the maximum value. This means that the curvature $\kappa $ at $p$ must be at least $1/R \ $ for any curve on $S$ passing through $p.$
This next part is what I don't understand
So every normal curvature is at least $1/R$ so $K(p)\geq 1/R^2 >0 $.
Why must every normal curvature be at least $1/R$ and why should that then mean $K(p)\geq 1/R^2 $ ? For the final part I suspect it is because the Gaussian curvature is the product of the principal curvatures, but why should they be at least $1/R$ ?
Because if there is a curve in $S$ through $p$ with less curvature, it wouldn't be able to stay inside the sphere centered at the origin with radius $R$. And it has to stay inside that sphere, as that sphere contains all of $S$.