I'm having trouble understanding the following situation. Apparently it's not difficult to see the union $V(x)\cup(\mathbb{A}^2\setminus V(y))$ is not a quasi-affine set.
Everything is being done over an algebraically closed field, and the $V(-)$ are the closed sets. So I think I have $$ V(x)=\{(a_1,a_2)\in\mathbb{A}^2:f(a_1,a_2)=0\} $$ where $f(x,y)=x$. So $V(x)=\{(0,b)\in\mathbb{A}^2\}$, the $y$-axis. Likewise $V(y)$ is the $x$-axis.
So I think $V(x)\cup(\mathbb{A}^2\setminus V(y))$ is geometrically the plane with the $x$-axis removed, except for the origin.
Is it somehow clear this is not quasi-affine? If this could be expressed as the set theoretic difference of closed sets, I believe it would have to be something like $\mathbb{A}^2\setminus Z$ where $Z$ is the $x$-axis except for the origin. It doesn't seem likely that there is a family of functions vanishing only there, is there a nice way to see it more rigourously?
Let $W := V(x) \cup (\mathbb{A}^2 \setminus V(y))$ be the set under consideration.
1) Show that the only closed subset of $\mathbb{A}^2$ containing $W$ is $\mathbb{A}^2$ itself.
2) Show that if a polynomial $f \in k[x,y]$ vanishes on $V(y) \setminus \{(0,0)\}$, then $f$ vanishes on $V(y)$ (on setting $x = 0$, $f(0,y)$ is a polynomial in $y$ with infinitely many roots).
Thus if $W$ were quasi-affine, then $W = \mathbb{A}^2 \setminus Z$ for some closed set $Z$. Now $Z$ is the vanishing locus of some polynomials which vanish on $V(y) \setminus \{(0,0)\}$, hence also on $V(y)$, which implies $(0,0) \in V(y) \subseteq Z$, contradicting $(0,0) \in W$.