= Solution
{c}
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:
$$
C^q=\prod_{i_0<\cdots<i_q} A,\qquad
(\delta s)_{i_0\cdots i_{q+1}}=\sum_{r=0}^{q+1}(-1)^r s_{i_0\cdots\widehat{i_r}\cdots i_{q+1}}.
$$
All restriction maps are identities. For $q\geq1$ define a <cochain homotopy> $h_q:C^q\to C^{q-1}$ by
$$
(h_qs)_{i_0\cdots i_{q-1}}=
\begin{cases}
0,&i_0=\alpha,\\
s_{\alpha i_0\cdots i_{q-1}},&i_0>\alpha.
\end{cases}
$$
We verify $\delta h_q+h_{q+1}\delta=\operatorname{id}_{C^q}$ for every $q\geq1$. Fix $I=(i_0<\cdots<i_q)$. If $i_0=\alpha$, then $(h_{q+1}\delta s)_I=0$. In $(\delta h_qs)_I$, 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 $i_0>\alpha$, then
$$
(\delta h_qs)_I=\sum_{r=0}^q(-1)^r s_{\alpha,I\setminus i_r},\qquad
(h_{q+1}\delta s)_I=(\delta s)_{\alpha,I}
=s_I-\sum_{r=0}^q(-1)^r s_{\alpha,I\setminus i_r}.
$$
Adding 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 $s=\delta(h_qs)$: every <Čech cocycle> in positive degree is a <Čech coboundary>. Therefore, using the permitted comparison with ordered <Čech cohomology>,
$$
\boxed{\check H^q_{\mathcal U}(X,\underline A)=0\qquad(q\geq1).}
$$
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.
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