A point has no higher sheaf cohomology

Every open subset of $X$ is $\mathrm{\varnothing }$ or $X$, and a sheaf $\mathcal{F}$ is determined by the group $\mathcal{F}(X)$ together with the structure maps to and from $\mathcal{F}(\mathrm{\varnothing })=0$ of [F2]: the sheaf condition over the two possible covers of $X$ is automatic, and over the empty cover of $\mathrm{\varnothing }$ it forces $\mathcal{F}(\mathrm{\varnothing })$ to be the one-element group by [F5]. Evaluating at $X$ is therefore a functor from $\mathrm{A}\mathrm{b}(X)$ to abelian groups which is fully faithful, since morphisms of sheaves are exactly the group homomorphisms of the section groups, and essentially surjective, since a group $A$ with $\mathcal{F}(\mathrm{\varnothing }):=0$ and $\mathcal{F}(X):=A$ and the only possible restriction maps defines a sheaf on $X$; hence evaluation at $X$ is an equivalence of categories, and under it $\mathrm{\Gamma }(X,-)$ becomes the identity functor of abelian groups, which is exact. This also shows that a sequence of sheaves on $X$ is exact precisely when the sequence of groups of sections over $X$ is exact.

By [step 1.1] the functor $\mathrm{\Gamma }(X,-)$ is exact, so the positive right derived functors relative to the supplied injective resolution datum vanish: $H^q(X,\mathcal{F})=R_I^q\mathrm{\Gamma }(X,\mathcal{F})=0$ for every $q>0$ by [F3], applied to the identity functor of $\mathrm{A}\mathrm{b}(X)$ through the equivalence. In degree zero, [F4] gives the canonical isomorphism $H^0(X,\mathcal{F})\cong \mathrm{\Gamma }(X,\mathcal{F})=\mathcal{F}(X)$, which under the equivalence of [step 1.1] is the identity of the group $\mathcal{F}(X)$. This proves both assertions. ∎