Consider the spaces $X=\left(\mathbb{C} P^{\infty} \times S^{1}\right) /\left(\left\{x_{0}\right\} \times S^{1}\right)$ and $Y=\mathbb{C} P^{\infty} \times S^{3}$. I want to show that they have isomorphic cohomology rings but that they are not homotopy equivalent.
The cohomology ring of $Y$ is straightforward enough: $$ H^*(Y;\mathbb{Z}/2)\cong H^*(\mathbb{C} P^\infty;\mathbb{Z}/2)\otimes_{\mathbb{Z}/2} H^*(S^3;\mathbb{Z}/2)\cong\mathbb{Z}/2[x]\otimes_{\mathbb{Z}/2}\mathbb{Z}/2[y]/(y^2)\cong \mathbb{Z}[x,y]/(y^2)$$ where $x$ is a generator of degree 2 and $y$ is a generator of degree 3.
For $X$ we consider the fibration sequence
$$ *\to X\to\mathbb{C}P^\infty$$
Question 1 is this indeed a fibration sequence?
The corresponding Serre spectral sequence has $E_2$ page given by $$ E_2^{ij}=\begin{cases}\mathbb{Z}/2,& j=0,i=2k\\ 0,&\text{otherwise}\end{cases}$$ so all the differentials are trivial and $E_2=E_\infty$. So $H^*(X;\mathbb{Z}/2)\cong\mathbb{Z}/2[x]$, but I don't think this is isomorphic to the cohomology ring of $Y$, so I went wrong somewhere.
Question 2 is the ring structure correct?
and finally, once I showed the two cohomology rings are isomorphic
Question 3 how can I show $X$ and $Y$ are not homotopy equivalent?
Edit: Based on @KentaS's comment, the fibration sequence is indeed incorrect. @Thomas's reference suggests looking at the homotopy fiber of $p\colon Ci\to\mathbb{C} P^\infty$ - the extension of the projection $\pi\colon S^1\times\mathbb{C}P^\infty\to\mathbb{C}P^\infty$ to the mapping cone of the inclusion $i\colon S^1\to\mathbb{C} P^\infty\times S^1$ - but I can't figure out how to compute it. The author claims that the homotopy fiber is homotopy equivalent to $S^3$, citing this reference, but I don't understand it.