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Suppose $S$ lies in the closed ball of radius $R$ centred at $c$. By compactness, the continuous function $q\mapsto |q-c|$ attains its maximum $r\le R$ at some $p\in S$. The sphere of radius $r$ about $c$ is a supporting sphere tangent to $S$ at $p$.

Put $N=(p-c)/r$. For any unit vector $v\in T_{p}S$, choose a surface curve $\alpha$ with $\alpha (0)=p$ and ${\alpha }^{\prime }(0)=v$. Since $|\alpha (t)-c{|}^2$ has a local maximum at zero,The scalar $⟨{\alpha }^{\prime \prime }(0),N⟩$ is the normal curvature in direction $v$, so every normal curvature is at most $-1/r$ with this choice of unit normal. Applying this to principal directions shows that both principal curvatures satisfy $k_i\le -1/r$. ThereforeThis proves both assertions, including the existence of an elliptic point on every compact regular surface.

Since $K\ge 0$, Gauss-Bonnet gives $4\pi (1-g)\ge 0$, hence $g\le 1$. 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 $g\ne 1$, and $g=0$. The classification theorem for surfaces now shows that $S$ is diffeomorphic to the sphere.

If a compact regular surface lies in a closed Euclidean ball of radius $R$, maximize distance from the ball's centre. At a maximizing point, comparison of the second fundamental form with the tangent supporting sphere makes both principal curvatures have magnitude at least $R^{-1}$ and the same sign. The Gaussian curvature there is therefore at least $R^{-2}$.