As the title shows, we have a fibration of $\mathbb{Z}_2\rightarrow E\mathbb{Z}_2\rightarrow B\mathbb{Z}_2\sim\mathbb{R}P^\infty$. I am trying to check my understanding of Serre spectral sequence with this simple example. First note that $H_i(\mathbb{Z}_2,\mathbb{Z})=\mathbb{Z}\oplus\mathbb{Z}$ for $i=0$ (since this fibre has two disconnected components) and otherwise $0$. By universal coefficient theorem , we get $H^i(\mathbb{Z}_2,\mathbb{Z})=\mathbb{Z}\oplus\mathbb{Z}$. We also know that $H^0(\mathbb{R}P^\infty,\mathbb{Z})=\mathbb{Z}$, $H^i(\mathbb{R}P^\infty,\mathbb{Z})=\mathbb{Z}_2$ with $i$ even and positive, and otherwise $0$. Therefore, the spectral sequence reads
\begin{array}{|c c c c c c} 0& 0 &0 & 0 & 0 & 0 & 0 & 0 \\ 0& 0 &0 & 0 & 0 & 0 & 0 & 0 \\ 0& 0 &0 & 0 & 0 & 0 & 0 & 0 \\ 0& 0 &0 & 0 & 0 & 0 & 0 & 0 \\ \mathbb{Z}\oplus\mathbb{Z}& 0 &\mathbb{Z}_2\oplus\mathbb{Z}_2 & 0 & \mathbb{Z}_2\oplus\mathbb{Z}_2 & 0 & \mathbb{Z}_2\oplus\mathbb{Z}_2 & 0 \\ \hline \end{array}
The differentials all seems to be trivial. It determines the cohomology groups of $E\mathbb{Z}_2$. However, the cohomology ring of $E\mathbb{Z}_2$ has to be trivial since it should be a (weakly) contractible space. There is a contradiction here. What's wrong with my understanding? Perhaps, Serre spectral sequence does not work for fibre disconnected or base space not simply-connected? Thanks in advance!