The shape operator $S$ defines a quadratic form on the tangent space $T_p(M)$, the second fundamental form $\langle Sv, v \rangle$. If $e_1, e_2$ are unit eigenvectors of $S$ associated to the eigenvalues $\kappa_1, \kappa_2$, then write $v = v_1 e_1 + v_2 e_2$. Then it's not hard to see that $\langle Sv, v \rangle = \kappa_1 v_1^2 + \kappa_2 v_2^2$. This implies that $\kappa_1$ is the minimum value of the second fundamental form as $v$ ranges over all unit vectors and $\kappa_2$ is the maximum value.
On the other hand, the normal curvature of a curve $\alpha : I \to M$ parameterized by arc-length at $\alpha(0) = p$ is precisely $\langle S \dot{\alpha}(0), \dot{\alpha}(0) \rangle$. Once you know that associated to every unit vector of the tangent space is a geodesic whose tangent vector is that vector, you're done.
This is a non-trivial fact. It is possible to create surfaces of revolution of the form
$$(\phi(v) \cos(u), \phi(v) \sin(u), \psi(v)),\quad \phi \ne 0$$
If $v$ is arc length we have $(\phi')^2 + (\psi')^2 = 1$. Having constant Gaussian curvature, K, then implies $\phi''+K\phi = 0$ and $\psi = \int \sqrt{1-(\psi')^2}\ dv$.
With $K=1$ and the curve intersection the X-Y plane perpendicularly we have
$$\phi(v) = C \cos v,\quad \psi(v) = \int_0^v \sqrt{1-C^2 \sin^2 v}\ dv.$$
So $\psi(v)$ is an Incomplete elliptic integral of the second kind. (Reference Do Carmo)
Now these surfaces have constant positive Gaussian curvature, if $C=1$, it gives a sphere, if $C\ne 1$, you have surface which have two singular points on the rotation axis. Examples of such surfaces can be seen at Wolfram demonstrations.
One of the comments above points to a looseness in Wikipedia's statement. The sphere is the only compact, simply-connected, Riemannian surface of constant positive curvature. The above examples, either fails to be compact if we exclude the singular points, or if they are left it fails to be smooth, hence not Riemannian.
Wikipedia does give a hint to a proof of the fact, it references the Liebmann's theorem (1900), and a brief comment that
A standard proof uses Hilbert's lemma that non-umbilical points of extreme principal curvature have non-positive Gaussian curvature.
This may have come from Hilbert and Cohn Vossen (p228). They first show that surface of constant positive Gaussian curvature, without boundary or singularities, must be a closed surface. Apart from the sphere there are no surfaces where both principal curvatures are constant. So we just need to consider cases like the above where the two principle curvatures vary, but their products are constant. As the surface is closed there must be a point where one of the principle curvature obtains its maximum value. But it can be proved analytically that such points can't exist, except on the boundary, on surfaces with constant positive Gaussian curvature.
do Carmo, Manfredo Perdigão, Differential geometry of curves and surfaces. Mineola, NY: Dover Publications (ISBN 978-0-486-80699-0/pbk). xvi, 510 p. (2016). ZBL1352.53002.
Best Answer
This is just a linear algebra fact.
The shape operator is known to be symmetric, so by the spectral theorem it has real eigenvalues, so there is an orthogonal basis in which this operator has a diagonal matrix, and the determinant of a diagonal matrix is easy to compute.
Of course, the validity of my "proof" depends on how you define the principal curvatures, but I rely here on the treatment given in the book of Theodore Shifrin "Differential Geometry. A first Course in Curves and Surfaces", where one can find a very accessible exposition.