why is the spectrum of the schrödinger operator discrete?
Solution 1:
I don't know the result in the case of $\mathrm L^2(\mathbb R^n)$, but the condition you describe : $\mathrm V(x) \to \infty$ for $||x|| \to \infty$ is exactly saying that $\mathrm V$ is a proper map (i.e the inverse image of any compact is a compact). A continuous map $\mathrm V$ on a compact manifold is automatically proper and maybe it is sufficient to adapt the theorem you state for $\mathbb R^n$.