= Solution
{c}
In an <irreducible topological space>, any two nonempty <open subsets> intersect. Let $W\subseteq X$ be a nonempty <open subset>. Two nonempty subsets open in $W$ are also open in $X$, and hence intersect; thus $W$ is itself an <irreducible topological space>. If $W$ were disconnected, a separation into two disjoint nonempty <open subsets> would contradict this intersection property. Hence $W$ is a <connected space>. The empty <open subset> is connected as well under the usual convention.
Discard empty members of $\mathcal U$ as in (b). Every finite intersection $U_{i_0}\cap\cdots\cap U_{i_q}$ is nonempty by repeated use of the intersection property and is an <open subset>, hence a <connected space>. A <locally constant function> on this intersection is constant. It follows that the <constant sheaf> has
$$
\underline A(U_{i_0}\cap\cdots\cap U_{i_q})\cong A,
$$
and every restriction map between these nonempty intersections is the identity on $A$. The whole <Čech cochain complex> is therefore the same as for the indexed <open cover> of a point having the same index set. The <cochain homotopy> constructed in (b) applies component by component and proves
$$
\boxed{\text{Every open subset of }X\text{ is connected},\qquad
\check H^q_{\mathcal U}(X,\underline A)=0\quad(q\geq1).}
$$
The argument concerns the <constant sheaf> of <locally constant functions>; arbitrary coefficient sheaves on an <irreducible topological space> need not have the same <Čech cohomology>.
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