Deriving the geodesic equation via the variational approach, you look for a curve at which the action is stationary, you do not check whether the stationary curve is maximum, minimum, or no extremum. If it is a maximum with one sign convention of the metric, it is a minimum with the opposite sign convention. So the issue "maximum or minimum" is not fundamental, but convention.

Whether the action has to be proportional to ds (why not some scalar function of ds?) is a different issue. Sometimes one can see expressions for (d\tau)^2 = ... or the -(d\tau)^2 = ..., with \tau the proper time; this again is convention. As different textbooks use different conventions and make corresponding statements, this may be confusing and one has to make sure to about the convention used in each text.

By what argument can it be said that the equation of motion of a massive particle can be obtained by finding a curve that renders the proper time between given endpoints stationary?

By what argument can it be said in classical Lagrangian mechanics that the equation of motion is obtained by finding a curve that renders the integral of T-V along the path stationary (in many cases, at least)? Because it turns out that the Euler-Lagrange equations for this variational problem yield the correct Newtonian equations of motion.

In general relativity, a particle only under the influence of gravity is force free, and, with U the 4-velocity, its equation of motion is \nabla_U U = 0, with \nabla_U the covariant derivative in direction U. This corresponds to the Newtonian "acceleration = 0" for a particle under no forces.

Writing \nabla_U U = 0 in components, with U^a = dx^a/d\tau, one obtains the component version of the geodesic equation (\nabla_U U = 0 is the component-free version). In a second step one can identify (or verify) the geodesic equation as the Euler-Lagrange equation of the Lagrangian L = g_{ab}(dx^a/d\tau)(dx^b/d\tau).