Covering a Riemannian manifold with geodesic balls without too much overlap I'm looking for a proof of the following fact:
Let $M$ be a compact Riemannian manifold. There is a natural number $h$, such that for any sufficiently small number $r>0$, there exists a cover of $M$ by geodesic balls of radius $r$ such that any $h$ of the balls have empty intersection.
 A: There is a neighborhood about each point where the exponential map is bilipschitz with constants arbitrarily close to 1. This implies that the images of balls of radius $R$ under the exponential map are almost balls of radius $R$, i.e. they contain balls of radius, say, $3R/4$ and lie in balls of radius $5R/4$. Cover your space with finitely many such neighborhoods.
Take a uniform cubical lattice in $R^n$ such that the $r$-neighborhood of the lattice points covers $R^n$ and send it to each of the chosen neighborhoods via the exponential map. The balls of radius $4r/3$ about the images of the lattice points cover the manifold and intersect only a bounded number of balls from points coming from the same exponential map. (Note the bound does not depend on r). But each point is only in finitely many images of the exponential map, so there is a fixed number on how many balls of radius $4r/3$ can contain any point, which gives a bound on the number that can intersect.
This is rough, but I feel confident in it, let me know what you think.
TL;DR the number of neighborhoods in the cover is finite by compactness, and each point only intersects a bounded number of balls in each cover because a Riemannian metric is Bilipschitz eq. to the Euclidean metric. 
