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?



Why Cech cohomology does not compute sheaf cohomology on an open annulus
Functions $(f\_0,f\_1,f\_2)$ as you requested do in fact exist. Your confusion comes from the fact that you are trying to impose the cocycle condition on the intersection $U\_0\cap U\_1\cap U\_2$, which is empty. That is the $f\_i$'s are such that
$$ f\_2(x) = f\_1(x)+1\ \forall x\in U\_1\cap U\_2$$
$$ f\_2(x) = f\_0(x)+2\ \forall x\in U\_0\cap U\_2$$
$$ f\_1(x) = f\_0(x)+3\ \forall x\in U\_0\cap U\_1$$
To get a contradiction you'd need an $x$ such that these three things simultaneously hold. But such an $x$ doesn't exists.


