Theorem 8.1 in textbook Differential Forms in Algebraic Topology (Bott and Tu) In theorem 8.1 the following statement is shown:
\begin{equation}
H_D(C^*(\mathfrak U,\Omega^*))\simeq H^*_\text{DR}(M),
\end{equation}
where the lefthand side is cohomology of a double complex and the righthand side  is the de Rham cohomology.
In this proof the map
\begin{equation}
r: \Omega^*(M) \rightarrow \Omega^*(U)\oplus\Omega^*(V)\subset C^*(\mathfrak U,\Omega^*),
\end{equation}
is considered and the commutativity of the following is used.
\begin{equation}
\begin{array}{ccc}
\Omega^*(M)& \overset{r}\longrightarrow& C^*(\mathfrak U,\Omega^*)\\
d\Big\uparrow& & \Big\uparrow D\\
\Omega^*(M)& \underset{r}\rightarrow& C^*(\mathfrak U,\Omega^*)
\end{array}
\end{equation}
The book says it is commutative because
\begin{align*}
Dr &= (\delta + (-1)^pd)r & [\because\text{it is the definition of $D$ and $p=0$}]\\
&= dr = rd.
\end{align*}
I don't understand why operator $\delta$ vanishes in the second line?
p.s.
The double complex $C^*(\mathfrak U,\Omega^*)$ is defined,
for open cover $\mathfrak U = \{U,V\}$, as
\begin{equation}
C^0(\mathfrak U,\Omega^q) = \Omega^q(U)\oplus\Omega^q(V),\;\;
C^1(\mathfrak U,\Omega^q) = \Omega^q(U\cap V).
\end{equation}
 A: Since the open cover has only two elements, we know that $\delta$ vanishes when $p>0$. But, regardless, $\delta\circ r = 0$ by definition, since $(r\omega)|_{U\cap V} = (r\omega)|_{U\cap V}$ for any $\omega\in\Omega^*(M)$.
EDIT: I had to go look at the book. I see that they haven't been explicit yet about defining the coboundary map on Čech cochains. They do in the paragraphs preceding this theorem say that $\delta$ is the difference map going horizontally. That is, if $\sigma = (\sigma_i)\in C^0(\mathfrak U,\Omega^q)$ is a $0$-cochain (i.e., an assignment of a smooth $q$-form to each open set $U_i$), then $\delta(\sigma)\in C^1(\mathfrak U,\Omega^q)$ is the $1$-cochain that assigns to the open set $U_i\cap U_j$ the $q$-form $\sigma_i\big|_{U_i\cap U_j} - \sigma_j\big|_{U_i\cap U_j}$. Thus, $\sigma$ is a $0$-cocycle if and only if $\sigma_i$ and $\sigma_j$ agree on the overlap $U_i\cap U_j$. This means that a $0$-cocycle gives a global form on $M$, and conversely.
A: I don't think you need to actually go into later discussions of Cech cohomology to get this result. The simpler answer is that the vanishing of $\delta r$ follows immediately from the exactness of the sequence:
$$ 0\rightarrow \Omega^*(U\cup V) \overset{r}\longrightarrow \Omega^*(U)\oplus \Omega^*(V) \overset{\delta}\longrightarrow \Omega^*(U\cap V) \rightarrow 0$$
If I recall correctly, the authors discuss this sequence a page or two before Theorem 8.1, and they prove the exactness of the above sequence (in a manner similar to Ted's answer) in Chapter 5.
