Proving the existence of good covers Usually one proves the existence of good covers in compact manifolds by Riemannian methods: we pick an arbitrary Riemannian metric, prove that geodesically convex neighborhoods exist, that they are closed under finite intersections, and diffeomorphic to balls; this is, for example, the argument that Bott and Tu sketch in their book.

Is there a non-Riemannian approach to this?

While this is not necessary for most things, it is a nice fact that good covers can be found which realize the covering dimension bound. 

Is there a differential-topological way to find them?

 A: The answer to both questions is yes.  Fix a triangulation of the manifold. For any vertex $v$ denote by $U_v$ the union of the relative interiors of all the faces of all dimensions that contain the vertex $v$. (Note: the vertex $v$ itself is a  face containing $v$  and it coincides with its relative interior.)  The set $U_v$ is open and contractible  and the resulting open cover is good. Its nerve is  is the simplicial set underlying the chosen triangulation.   This cover answers both your questions. 
A: I think you can obtain a good cover of $C^2$ manifold (compact or not) from the charts/atlas definition and a little bit of topology (locally finite atlas and a relatively compact "shrinking" of it).
The very simple idea (akin to that in Vitali's answer) is that under a $C^2$ diffeomorphism between open subsets of euclidean $n$-space, the (pre-)image of a sufficiently small ball centered at a point will be convex, as soon as the curvature of its boundary "dominates" the second derivative of the diffeomorphism (or its inverse).
In formulas, if $\phi$ is the diffeomorphism, this boils down to the fact that the $C^2$ function $x\mapsto |\phi(x)-\phi(x_0)|^2$ has a positive definite hessian at $x_0$, hence is convex near $x_0$. 
With a little more care, I think you can still conclude if $\phi$ is only $C^{1+Lip}$. 
A: you don't really need a whole lot of Riemannian geometry to prove this. Embed the manifold into $\mathbb R^n$ by Whitney and look at very small charts around points given by orthogonal projections onto the tangent spaces. the transition maps will be arbitrary close to identity in $C^2$. that means that a small round disk in one chart will remain strictly convex in nearby charts (because if $f(x)=|x|^2$ and $\phi$ is a transition map  such that $\phi-Id$ has small first and second derivatives then $f\circ \phi$ is still strictly convex and hence has convex sublevel sets). This is is all you need to conclude that all intersections are contractible. I guess since the above argument doesn't use any Riemannian geometry notions it should qualify as an answer to  the second question?
Incidentally, does a good open cover always exist if a manifold is only topological?
