I'm going through Čech cohomology at a gentle pace and as you know we need a colimit to obtain the sought-after sheaf cohomology. In practice, for calculating actual cohomology groups, I would invoke Leray's theorem and choose an acyclic covering, so that there is no need for a colimit, and that's quite clear so far. On sunny days I also have a version of Mayer-Vietoris that makes computations tractable.
My issue is that I assumed quite naively that contractible spaces (say, open disks) have trivial cohomology (say, for the constant presheaf). And I get the expected results so it's not obviously wrong. But how can I prove this?
I'm aware that Čech cohomology is isomorphic to other cohomology theories for "nice enough" space, and that would be a way (albeit an impractical one) to get a proof. But surely there is a simple and elegant argument "internal" to Čech cohomology?
Čech cohomology functor is homotopy invariant. Meaning homotopic maps are mapped to equal homomorphisms. This implies that a homotopy equivalence $f:X\to Y$ induces an isomorphism
$$f^k: H^k(Y,\mathscr{F}) \to H^k(X, f^{-1}\mathscr{F})$$
for any $k$ and any "suitable" (i.e. locally constant) sheaf $\mathscr{F}$. Here $f^{-1}\mathscr{F}$ means the inverse image of a presheaf which in case of a constant map (i.e. the contractible case) is a constant sheaf associated with $\mathscr{F}(Y)$. For more details (including proofs) see this paper.
So to calculate Čech cohomology of a contractible space all you have to do is to calculate Čech cohomology of a point. Which can be done straight from the definition. That's because every presheaf is trivial (meaning uniquely determined by a single abelian group) and there is only one open cover containing only one open set. It requires some effort but all calculations are very doable and I encourage you to try it.