The Stacks project

Proof. Choose an injective resolution $\mathcal{F}\to \mathcal{I}^\bullet $ in $\textit{Ab}(\mathcal{C})$, and consider the double complex $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}^\bullet )$ and the maps

Here the horizontal map is the natural map $\Gamma (U, I^\bullet ) \to \check{\mathcal{C}}^0(\mathcal{U}, \mathcal{I}^\bullet )$ to the left column, and the vertical map is induced by $\mathcal{F}\to \mathcal{I}^0$ and lands in the bottom row. By assumption, $\mathcal{I}^\bullet $ is a complex of injectives in $\textit{Ab}(\mathcal{C})$, hence by Lemma 59.19.1, it is a complex of injectives in $\textit{PAb}(\mathcal{C})$. Thus, the rows of the double complex are exact in positive degrees (Lemma 59.18.7), and the kernel of $\check{\mathcal{C}}^0(\mathcal{U}, \mathcal{I}^\bullet ) \to \check{\mathcal{C}}^1(\mathcal{U}, \mathcal{I}^\bullet )$ is equal to $\Gamma (U, \mathcal{I}^\bullet )$, since $\mathcal{I}^\bullet $ is a complex of sheaves. In particular, the cohomology of the total complex is the standard cohomology of the global sections functor $H^0(U, \mathcal{F})$.

For the vertical direction, the $q$th cohomology group of the $p$th column is

in the entry $E_1^{p, q}$. So this is a standard double complex spectral sequence, and the $E_2$-page is as prescribed. For more details see Cohomology on Sites, Lemma 21.10.6. $\square$