This question will be a bit long because its subject is quite technical.
First some preliminaries: If you have a cover of a space that is open or locally finite and closed you can try to compute sheaf cohomology of the space from sheaf cohomology of elements of the cover.
One well-known instance of this approach is Leray's theorem:
Theorem
Assume that $H^{i>0}(U_i;\mathscr{F})=0$ for all elements of the cover $\mathscr{U}=\{U_i\}_{i\in I}$ of the space $X.$ Then there is an isomorphism $$ H^*(X;\mathscr{F})\cong H^*_{\mathscr{U}}(\mathscr{F}). $$
There is, however, a finer result which implies Leray's theorem.
Theorem
Let $\mathscr{U}=\{U_i\}_{i\in I}$ be a cover of the space $X.$ Then there is an isomorphism $$ H^*(X;\mathscr{F})\cong H^*_{\mathrm{Tot}}(C^*(\mathscr{U};\mathscr{G}^*)). $$ Here $\mathscr{G}^*$ is the Godement resolution of the sheaf $\mathscr{F}$ (actually any flasque resolution will do).
Now standard methods for the cohomology of the double complex imply the following result:
Theorem
There is a spectral sequence with converging to $H^*(X;\mathscr{F})$ induced by the column filtration. Its $E_1$ page is isomorphic to
$$ E_1=C^*(\mathscr{U};\mathscr{H}_\mathscr{F}^*) $$ Here $\mathscr{H}_\mathscr{F}^*$ is a local system on the nerve of $\mathscr{U}$ given by $\mathscr{F}$-cohomology of elements $U_I.$
Now, what I was trying to do. I tried to compute sheaf cohomology of a space using the spectral sequence above. In my case nerve of a cover is just a 2-simplex.
My goal was to prove that the spectral sequence collapses (not always but in my particular case) at the $E_2$-page. To do this I have used an explicit computation with double complex, which forms the page $E_0$ of the spectral sequence. As a resolution for my sheaf, I used Godement resolution.
Now what I have obtained is clearly wrong, because it seems that the spectral sequence always collapses at the term $E_2$ (without any additional assumptions).
The computation goes like this. Denote the vertices of the nerve by $0,1,2.$ Denote the sheaf for which we are trying to compute by $\mathscr{F}.$ Denote by $\mathscr{G}^i$ the $i$-th term of the Godement resolution for the sheaf $\mathscr{F}.$ And let elements of the cover be denoted by $U_I,$ where $I$ is a set with elements in $\{0,1,2\}.$
Then a generic element of the group $E_0^{0,1}$ looks like $\xi=(\xi_0,\xi_1,\xi_2).$ Here $\xi_i\in \mathscr{G}^1(U_i).$ Now assume that $\xi$ is a cycle with respect to $d_1.$ Then $d_{\check{C}}(\xi)=d_{G}\nu.$ Here $d_G$ is the differential in Godement resolution and $d_{\check{C}}$ is the Čech differential. The element $\nu$ is in $E_0^{1,0}.$
Explicitly the equation $d_{\check{C}}(\xi)=d_{G}\nu$ means that
$$ ((\xi_1-\xi_0)\vert_{U_{01}},(\xi_2-\xi_1)\vert_{U_{12}},(\xi_2-\xi_0)\vert_{U_{20}})=\left(\prod_{x\in U_{01}} \nu^{01}_x, \prod_{y\in U_{12}} \nu^{12}_y,\prod_{z\in U_{20}} \nu^{20}_x\right)=d_G \nu $$
Now to prove that $d_2(\xi)=0$ we need to check that $d_{\check{C}}(\nu)=0.$ By definition we have:
$$ d_{\check{C}}(\nu)=\prod_{x\in U_{012}} (\nu^{12}_x-\nu^{20}_x+\nu^{01}_x)=(\xi_1-\xi_0)\vert_{U_{012}}+(\xi_2-\xi_1)\vert_{U_{012}}-(\xi_2-\xi_0)\vert_{U_{012}}=0\vert_{U_{012}} $$
Now to prove that the whole $d_1$ is trivial I just need to replace the sheaf $\mathscr{F}$ with an appropriate term of the Godement resolution. Then I just repeat the argument above.
What am I getting wrong? It seems that by using similar computations we can prove that the only non-trivial differential in this spectral sequence is always $d_1.$ This is obviously not true.