Chart gluing: a set closed in all four charts and containing the affine curve $P_\alpha=0$ contains $V_\alpha$

Let $m\ge1$ be an integer and let $\alpha\in\mathbb C$ be arbitrary. Let $P_\alpha(X,Y)=X^m(\alpha+XY)+Y^m(\bar\alpha+XY)$ be the polynomial of equation (6), and let $V_\alpha\subseteq\mathbb P^1\times\mathbb P^1$ be the zero locus of its bihomogenization of bidegree $(m+1,m+1)$ in homogeneous coordinates $([x_0:x_1],[y_0:y_1])$,

so that $F_\alpha((X,1),(Y,1))=P_\alpha(X,Y)$. Consider the four standard charts $\phi_i\colon\mathbb C^2\to\mathbb P^1\times\mathbb P^1$, $\phi_0(X,Y)=([X:1],[Y:1])$, $\phi_1(u,Y)=([1:u],[Y:1])$, $\phi_2(v,X)=([X:1],[1:v])$ and $\phi_3(u,v)=([1:u],[1:v])$, and give $\mathbb C^2$ its Zariski topology, whose closed sets are the common zero sets of families of polynomials in two variables. Let $S\subseteq\mathbb P^1\times\mathbb P^1$ be a set such that $\phi_i^{-1}(S)$ is Zariski closed in $\mathbb C^2$ for each $i=0,1,2,3$.

Then, if $S$ contains the affine curve in the chart $\phi_0$, it contains all of $V_\alpha$:

This is the gluing step in identifying $V_\alpha$ with the closure of the affine curve (6) in $\mathbb P^1\times\mathbb P^1$, the curve whose singular points and genus Lemma 4 describes: a set that is closed in every chart and contains the affine curve already contains all of $V_\alpha$, boundary points included.

Formalization Note: no topology on $\mathbb P^1\times\mathbb P^1$ is used; closedness of $S$ is required chart by chart, for the Zariski topology on $\mathbb C^2$ induced from the prime spectrum through evaluation ideals (a local instance). $V_\alpha$ is defined by evaluating $F_\alpha$ at chosen representatives of the two points, which does not depend on the choice by bihomogeneity.