Let $$S^2 = \{(x,y,z)\in\mathbb{R}^3 : x^2 + y^2 + z^2 = 2\}$$ and $\tilde{X}=S^2\times S^2\subset\mathbb{R}^6$. Define an equivalent relation $\sim$ on $S^2\times S^2$ setting $$(x_1,y_1,z_1,x_2,y_2,z_2) \sim (-x_1,-y_1,-z_1,-x_2,-y_2,-z_2).$$
I want to show that the quotient space $X = \tilde{X}/\sim$ is not homeomorphic to the cartesian product of the real projective plane with itself.
We can see $X$ as the orbit space of $\tilde{X}$ under the group action of the group $G = \{\text{id}_\tilde{X}, a\}$, where $a : \tilde{X} \to \tilde{X}$ is multiplication by $-1$. Since this is a properly discontinuous action (at least it seems so to me), and since $S^2$ is simply connected, we must have $$\pi_1(X) \cong G\cong \mathbb{Z}_2,$$ which is not isomorphic to $$\pi_1(\mathbb{R}\mathbb{P}^2\times \mathbb{R}\mathbb{P}^2) \cong \mathbb{Z}_2\times \mathbb{Z}_2.$$
It seems weird to me that $X$ would have the same fundamental group as $\mathbb{R}\mathbb{P}^2$ and I am not confident this reasoning is correct. I would appreciate any feedback.