 Solution

Discard empty members of the open cover: every Čech cochain group factor supported on an empty intersection is the zero abelian group, so this changes no Čech cohomology. Every remaining member is the one-point space $X$; the index set is nonempty even if the same open subset is listed repeatedly. Choose a well-order with least index $\alpha$. Use the ordered Čech cochain complex permitted in the question:All restriction maps are identities. For $q\ge 1$ define a cochain homotopy $h_q:C^q\to C^{q-1}$ byWe verify $\delta h_q+h_{q+1}\delta ={\mathrm{id}}_{C^q}$ for every $q\ge 1$. Fix $I=(i_0<\cdots <i_q)$. If $i_0=\alpha$, then $(h_{q+1}\delta s{)}_I=0$. In $(\delta h_{q}s{)}_I$, the term deleting $i_0$ is $s_I$ and every other term is zero because its retained first index is $\alpha$. Thus the identity holds on $I$.

If $i_0>\alpha$, thenAdding proves the identity in this case also. Each formula uses only finitely many terms on each component, so it remains valid for an infinite open cover and arbitrary abelian group $A$.

If $s\in C^q$ is a Čech cocycle, then $\delta s=0$, and the identity gives $s=\delta (h_{q}s)$: every Čech cocycle in positive degree is a Čech coboundary. Therefore, using the permitted comparison with ordered Čech cohomology,The identity is asserted in positive degrees; degree zero has $H^0\cong A$, so an unaugmented contraction to zero in every degree would be incorrect.