Let $k$ be any field. If $P_1,P_2\subset k[x_1,\dots, x_n]$ are subsets of polynomials, then I want to show that $$Z(P_1P_2)=Z(P_1)\cup Z(P_2)$$ where $Z(P)$ is a vanishing set of $P\subset k[x_1,\dots, x_n]$ i.e. $$Z(P)=\{a\in\mathbb{A}^n:f(a)=0\text{ for all }f\in P\}$$ and $P_1P_2=\{fg:f\in P_1,g\in P_2\}$. I want to check that my proof is correct.
Proof: The inclusion $Z(P_1)\cup Z(P_2)\subseteq Z(P_1P_2)$ is obvious since if $a\in Z(P_1)\cup Z(P_2)$ then $a\in Z(P_1)$ or $a\in Z(P_2)$. Assume that $a\in Z(P_1)$, then $f(a)=0$ for all $f\in P_1$ which gives that $(fg)(a)=0$ for all $f\in P_1$ and $g\in P_2$ i.e. $a\in Z(P_1P_2)$.
The second inclusion $Z(P_1P_2)\subseteq Z(P_1)\cup Z(P_2)$ can be proved using a contrapositive proof. But, I want to check the my direct proof is correct. If $a\in Z(P_1P_2)$, then $(fg)(a)=0$ for all $f\in P_1$ and $g\in P_2$. If we can find $g\in P_2$ s.t. $g(a)\neq0$, then we can see that $a\in Z(P_1$ as $f(a)g(a)=(fg)(a)=0$ forces $f(a)=0$ for all $f\in Z(P_1)$. If we cannot find such $g\in P_2$, then it means that $g(a)=0$ for all $g\in S_2$ i.e. $a\in Z(P_2)$.
Yes, your proof is fine. I would draw attention to/you should be aware of the fact you have used a special property where you have said "$f(a)g(a)=(fg)(a)=0$ forces $f(a)=0$", that is, that $f(a)g(a)$ lies in a field and thus an integral domain. This is what guarantees the same fact to still hold when we generalise to the prime spectrum....