Construction of $\mathbb{CP}^2$ from $S^2\times S^2$

Show that $\mathbb{CP}^2$ is homeomorphic to the orbit space of $(S^2\times S^2)/\mathbb{Z}_2$. The action is given by $(x,y) \mapsto (y,x)$.

My attempt: I can see that $S^2\times S^2$ is a subset of $S^5$. I am unsuccessfully trying to restrict the action of $S^1$ on $S^5$ to $S^2\times S^2$ to get the required map.

This is actually a special case of a more general statement involving the symmetric products of $S^2$. The $n$-fold symmetric product $SP_n(S^2)$ is the quotient space $$SP_n(S^2)=(S^2)^n/\Sigma_n$$ where $(S^2)^n$ denotes the $n$-fold cartesian product, and $\Sigma_n$, the group of permutations on $n$ letters, acts by permuting the factors. The claim is that for each $n\in\mathbb{N}$ there is a homeomorphism $$SP_n(S^2)\cong \mathbb{C}P^n.$$ Your question relates to the special case of $n=2$, but it's no more trouble to establish the general result.

To begin introduce $P_n(\mathbb{C})$ as the set of non-zero complex polynomials of degree $\leq n$. The assignment $$P_n(\mathbb{C})\ni\sum^n_{i=1}a_iz^i\mapsto (a_0,\dots,a_n)\in\mathbb{C}^n\setminus\{0\}$$ establishes a bijection between $P_n(\mathbb{C})$ and $\mathbb{C}^n\setminus\{0\}$ which we will use to topologise $P_n(\mathbb{C})$. This yields the quotient map $$\gamma_n:P_n(\mathbb{C})\rightarrow \mathbb{C}P^n,\qquad \sum^n_{i=1}a_iz^i\mapsto [a_0,\dots,a_n].$$ Observe that if $f,g\in P_n(\mathbb{C})$, then $\gamma_n(f)=\gamma_n(g)$ if and only if $f$ and $g$ have the same roots (counted with multiplicity).

Now identify $S^2\cong\mathbb{C}P^1\cong\mathbb{C}\cup\infty$ as the Riemann plane and define $\varphi_n:SP_n(\mathbb{C})\rightarrow P_n(\mathbb{C})$ by $$\varphi_n\langle\xi_1,\xi_2,\dots,\xi_n\rangle=(z-\xi_1)(z-\xi_2)\cdots(z-\xi_n),$$ with the understanding to replace $(z-\infty)$ by $1$ if it appears in the product on the right-hand side. I am not claiming that $\varphi_n$ is continuous, but rather than its composite with $\gamma_n$ is a continuous map $SP_n(S^2)\rightarrow\mathbb{C}P^n$. You should convince yourself that this is true, and that $(z-\infty)\mapsto 1$ is indeed the correct replacement to make.

Now, we remarked above that the quotient $\gamma_n$ identifies two polynomials exactly when they have identical roots, so it follows that the composite $\gamma_n\varphi_n:SP_n(S^2)\rightarrow \mathbb{C}P^n$ is injective. On the other hand, to see that it is surjective, for $\xi_1,\dots,\xi_n\in\mathbb{C}\cup\{\infty\}$ we can write $$\prod^n_{i=1}(z-\xi_i)=\sum^n_{i=0}(-1)^i\left(\sum_{1\leq j_1<j_2<\dots< j_i\leq n}\xi_{j_1}\dots\xi_{j_i}\right) z^{n-i}$$ and so check that everything in the target indeed gets hit.

Finally we have that $\gamma_n\varphi_n$ is a continuous bijection. Since $SP_n(S^2)$ is compact and $\mathbb{C}P^n$ is Hausdorff we see that this map is a homeomorphism.

Edit: Thanks to Olivier Bégassat for a correction.