$D:\tilde{M}\rightarrow \Omega$ is a local isometry (developing map) onto a connected manifold $\Omega$. $\tilde{M}$ is simply connected, has constant sectional curvature and there exists a deck group $\pi(M)$ on $\tilde{M}$.
$\tilde{M}$ is the universal cover of a compact manifold $M$. The inverse set $D^{-1}(x)$ can be shown to be finite.
I want to show that $D$ is in fact a covering map, and to do so I would have to show that every inverse set $D^{-1}(x)$ has the same cardinality, i.e. the same number of finite points.
Is there a counterexample to this, or how could this be done?
I disagree with the claim that $D^{-1}(x)$ can be shown to be finite. Why not take the covering map $\mathbb{R}^2 \to S^1 \times S^1$? (And put the flat metric on the torus so it this map is a local isometry, and satisfies your curvature conditions.)
What is your condition about the deck group? That is not clear.
If you assume that $\tilde{M}$ is compact this result is easier. In that case, the hint would be: find a neighborhood for each point in the fiber which is a local isometry. Try to refine these neighborhoods to produce local trivializations. (A proper local diffeomorphism between connected manifolds is a covering map. Really this is a topological fact: When is a local homeomorphism a covering map?)
Maybe you should look in Do Carmo, in the section on space forms. He has the following theorem, which is related in spirit to your question:
(Thm. 4.1 in chapter 8) Let $M^n$ be a complete Riemannian manifold with constant sectional curvature $K$. Then the universal covering, with the coverin gmetric, is isoemetric to hyperbolic space, Euclidean space, or $S^n$, with their natural constant curvature metrics, and curvatures $-1$, $0$ and $1$ respectively.