Let $G$ be a group equipped with a left-invariant metric $d$: that is, $(G,d)$ is a metric space and $d(xy,xz) = d(y,z)$ for all $x,y,z \in G$. Suppose further that $(G,d)$ is connected, locally compact, and bounded. Must $(G,d)$ be compact?
I can show from the local compactness that $(G,d)$ is complete. I don't see how to get it to be totally bounded, but I also can't think of a counterexample.
If it helps, you can assume that $(G,d)$ is a topological group (i.e. right translation and inversion are homeomorphisms; we already know that left translation is an isometry). I would even be interested in the case where $G$ is a finite-dimensional Lie group and $d$ induces the manifold topology, but I do not want to assume that $d$ is induced by a left-invariant Riemannian metric.
The answer is no. Consider $\mathbb R$ with the bounded left invariant metric $$d(x,y)=\min(|x-y|,1)$$