This is Exercise 5.2.6 in Qing Liu's book Algebraic Geometry and Arithmetic Curve. In part (b), I can show (b) if the open covering has only finitely many open subsets, since the the colimit of the Cech complexes $C(\mathcal{U},\mathcal{F}_\lambda)$ would be the Cech complex of $\mathcal{F}$, $C(\mathcal{U},\mathcal{F})$. Then part (c) follows from the fact that finite open coverings are cofinal in the system of open coverings. So it is reasonable for me to question that if part (b) is true if $\mathcal{U}$ is not finite.
Clearly every $\mathcal{U}$ has a finite subcover $\mathcal{V}$, do we have that $$H^p(C(\mathcal{U},\mathcal{F}))\stackrel{\sim}{\to} H^p (C(\mathcal{V},\mathcal{F}))$$ for all $p\geq 0$ ? If so, we can reduce part (b) to the finite case.
Update: Nope, the above map is not an isomorphism in general, I asked it (see here). In particular, if $\mathcal{V}=\mathcal{U}\backslash {U_0}$, then we can calculate $\ker(H^{p+1}(\mathcal{U},\mathcal{F})\to H^{p+1}(\mathcal{V},\mathcal{F}))=H^p(\mathcal{V}\cap U_0,\mathcal{F})$ from a short exact sequence of complexes. So if we pick $U_0=X,p=0,\mathcal{F}=\underline{\mathbb{Z}}$ then $H^0(\mathcal{V}\cap U_0,\mathcal{F})=H^0(\mathcal{V},\mathcal{F})=\mathcal{F}(X)=\underline{\mathbb{Z}}(X)\neq 0$. Thus the above map is almost never an isomorphism unless $\mathcal{U}$ and $\mathcal{V}$ is carefully chosen (e.g. affine).