Let $G$ be a locally compact topological group (in all the cases I have seen so far it is compact) and $(X, d)$ a metric space. Suppose that $G$ acts on $X$ and for each $g \in G$, $x \mapsto g \cdot x$ is an isometry and for each $x \in X$, the orbit $o_x$ is closed.
Can I show that $$d^*(o_x, o_y) = \inf_{\substack{a \in o_x \\ b \in o_y}} d(a,b)$$ is a metric on $X/G$?
I showed that the isometry property implies that $d^*(o_x, o_y) = \inf\limits_{g \in G} d(g \cdot x, y)$ and I showed that if the orbits are closed, then $d^*(o_x, o_y) = 0$ implies that $o_x = o_y$. I haven't been able to show the triangle inequality, but I found this question: Quotient metric on a complete Riemannian manifold which made me think it might be possible under some extra assumptions since it seems to be possible in other spaces.
Edit: I just saw this question: Triangle inequality for the distance between two sets
and realized that \begin{align*} d(o_x, z) &= \inf\limits_{g \in G} d(g \cdot x, z)\\ &= d^*(o_x, o_z) \end{align*} So that \begin{align*} d^*(o_x, o_y) &\leq d(o_x, z) + d(z, o_y)\\ &= d^*(o_x, o_z) + d^*(o_z, o_y) \end{align*} I think that shows it, please correct me if I messed up.