Shortest path on a sphere
Here's a geometric observation that can hardly be called a "proof", but may be appealing nonetheless.
If $p$ and $q$ are distinct points of the sphere $S^{n}$, if $C:[0, 1] \to S^{n}$ is a "shortest path" joining $p$ to $q$, and if $F:S^{n} \to S^{n}$ is a distance-preserving map fixing $p$ and $q$, then $F \circ C$ is also a shortest path (because the length of $F \circ C$ is equal to the length of $C$).
Assume $q \neq -p$. If you believe there exists a unique shortest path from $p$ to $q$, it's not difficult to see that the "short" great circle arc is the only candidate: Every point not on the great circle through $p$ and $q$ is moved by some isometry of the sphere that fixes $p$ and $q$.
If you're thinking specifically of $S^{2}$, reflection $F$ in the plane containing $p$, $q$, and the center of the sphere is an isometry, and $f(x) = x$ if and only if $x$ lies on the great circle through $p$ and $q$.
(A similar argument "justifies" that the shortest path between distinct points of the Euclidean plane is the line segment joining them.)
You can show that great circle arcs are geodesics by parameterizing such an arc so that it has unit speed, and then showing that the acceleration along the arc is perpendicular to the sphere surface. (This assumes you accept that geodesics have the characteristic that their acceleration is perpendicular to the tangent plane of the surface at each point of the geodesic.)
Then uniqueness of the geodesic from a point in a direction shows it must be the great circle arc.
I realize this might not be what you seek because it doesn't connect directly to shortest paths...
By symmetry!
Simple geometric proof:
(parallel symmetry) Consider the plane that is the perpendicular bisector of the straight line segment joining the two points. All objects (i.e., the sphere and the points) are symmetric with respect to that plane, so if the path is unique, it must stay the same after it is reflected across the plane.
Why? Because otherwise it wouldn't be unique -- by reflecting the problem, we would be keeping the inputs the same, but changing the output, and hence the path wouldn't be a function of the inputs.(perpendicular symmetry) Now consider the plane of the great circle -- that is, the plane that goes through the center of the sphere as well as the two given points. Again, the sphere and the points are reflectively symmetric with respect to this plane, so if the path is unique, it must stay the same after being reflected across this plane. (Same reason as above.)
(spherical constraint) The path must, by definition of the problem, lie on the sphere.
(uniqueness) The path must be unique. (This is intuitively obvious, so I won't try to prove it.)
It's easy to see that the only path that satisfies these three conditions is the one on the great circle.
Why? Because the intersection of the sphere with the two planes of symmetry clearly satisfies conditions 1-3. Furthermore, condition 4 implies that no other path can be the shortest path. Hence, the intersection of these shapes must be the shortest path itself.
However...
This method only works for this problem. By contrast, calculus-based methods (see the Calculus of Variations) work for other problems that lack such symmetries, and hence you should still learn those approaches so you can solve the shortest path problem for e.g. ellipsoids.