Point whose neighbourhood is contained in a certain sphere of radius r has Gaussian curvature greater than $\frac{1}{r^2}$?

Question

I am doing a question which asks one to show that is a compact, regular surface $\Sigma$ has Gaussian curvature $K < \frac{1}{r^2}$, then $\text{diam}(\Sigma) > r$. The given solution is the following:

I don't understand how they concluded that $K(q) \geq s^{-2}$ if $\Sigma$ is contained in $\overline{B}_s(p)$. Intuitively this is clear, but I can't seem to find a theorem that states something like this explicitly. Where can I find such a result?