G-spaces and manifolds
On a *complete*smooth Riemannian manifold,
Any bounded (with respect to the distance function induced by the Riemannian metric) closed set in a manifold is compact.
This is telling you that there is a minimal geodesic joining $x$ to $y$ that, when extended, is also a minimal geodesic joining $x$ to $z_1$. And there is another minimal geodesic joining $x$ to $y$ that when extended is a minimal geodesic joining $x$ to $z_2$. But if there are two distinct geodesics joining $x$ to $y$, neither is minimal beyond $y$. So the two geodesics have to be the same and therefore $z_1 = z_2$.
CORRECTION: "complete" added to assumption above.
For a smooth manifold, you need to construct a distance function to get a G-space. One way to do this is to construct a complete Riemannian metric. I'm not certain that this can be done, but offhand if you take a locally finite covering by open sets diffeomorphic to the Euclidean ball, use the standard Euclidean metric on each ball (where each ball has radius $1$), and use a partition of unity subordinate to this cover to glue together these metrics, it seems to me that the resulting metric is complete.
For a topological manifold, I don't know.
If I am not wrong, I believe that every topological manifold admits a complete metric. With this metric, it is possible to give a Path Metric Space structure (in the sense of Gromov, see the book Metric Structures for Riemannian and non-Riemannian spaces). I guess that this structure allows to make the same arguments as for Riemannian manifolds for topological manifolds.
Anyway, I recomend this survey about Busemann conjecture which also discusses a stronger conjecture (the Bing-Borsuk conjecture).