Let $A=\{z\in\mathbf{C}:1/2<|z|<1\}$ be an open annulus. Let us cover $A$ by 3 open sets: $U_0,U_1$ and $U_2$ which we assume to be all homeomorphic to a 2 dimensional open disc. Moreover, we assume that $U_{ij}:=U_{i}\cap U_j$ are also homeomorphic to two dimensional open discs and that $U_0\cap U_1\cap U_2=\emptyset$. Let $\mathfrak{U}=\{U_0,U_1,U_2\}$.

Let $\mathcal{O}_A$ be the structure sheaf of $A$ where we think of $A$ as a complex analytic manifold. Since $A$, $U_i$'s and $U_{ij}$'s are connected open Riemann surfaces they are Stein manifolds and thus $H^q(A,\mathcal{O}_A)$, $H^q(U_i,\mathcal{O}_A|_{U_i})$ and $H^q(U_{ij},\mathcal{O}_A|_{U_{ij}})$ all vanish for $q\geq 1$. Therefore I would expect $$ \check{H}^{1}(\mathfrak{U},\mathcal{O}_A)\simeq H^1(A,\mathcal{O}_A)=0. $$ Since $$ C^1(\mathfrak{U},\mathcal{O}_A)=\mathcal{O}_A(U_{12})\times \mathcal{O}_A(U_{02})\times \mathcal{O}_A(U_{01}). $$ I would expect every 1-cochain to come from a 0-cochain. So for example, one could take a triple of constant functions (which are clearly holomorphic) for example $$ (1,2,3)\in C^1(\mathfrak{U},\mathcal{O}_A) $$ Thus there should exist $$ (f_0,f_1,f_2)\in \mathcal{O}_A(U_{0})\times \mathcal{O}_A(U_{1})\times \mathcal{O}_A(U_{2}) $$ such that $df=((f_2-f_1)|_{U_{12}},(f_2-f_0)|_{U_{02}},(f_1-f_0)|_{U_{01}})=(1,2,3)$. But this is clearly impossible since $2-1\neq 3$!

**Q:** So why is Chech cohomology not computing $H^1(A,\mathcal{O}_A)$ here?