Proof of The Product Topology is a Topology

For subsets $U$ of $X$ and $V$ of $Y$, every element of $U\times V$ is an ordered pair $(u,v)$ with $u\in U$ and $v\in V$ by The Cartesian Product of Two Classes §product, and $(u,v)\in U\times V$ if and only if $u\in U$ and $v\in V$ by Membership in a Cartesian Product, and the Cartesian Product of Two Sets Is a Set §membership. Membership in a union $\bigcup\mathcal{S}$ and in a power set is as in The Union Set and the Power Set of a Set §union and The Union Set and the Power Set of a Set §power. $X\times Y$ is a set by Membership in a Cartesian Product, and the Cartesian Product of Two Sets Is a Set §set, and by its definition $\mathcal{T}_{X\times Y}$ is a subset of $\mathcal{P}(X\times Y)$, as Topological Space §topology requires; we check the conditions Topological Space §empty-whole, Topological Space §unions and Topological Space §intersections.

Empty set and whole set. The subset $\emptyset$ of $X\times Y$ has no elements, so the condition defining $\mathcal{T}_{X\times Y}$ holds for it vacuously and $\emptyset\in\mathcal{T}_{X\times Y}$. By Topological Space §empty-whole, $X\in\mathcal{T}_{X}$ and $Y\in\mathcal{T}_{Y}$; for all $x\in X$ and $y\in Y$ with $(x,y)\in X\times Y$ the sets $U=X$ and $V=Y$ satisfy $x\in U$, $y\in V$ and $U\times V\subseteq X\times Y$. Hence $X\times Y\in\mathcal{T}_{X\times Y}$.

Unions. Let $\mathcal{S}$ be a subset of $\mathcal{T}_{X\times Y}$. Every element of $\bigcup\mathcal{S}$ lies in some $W\in\mathcal{S}$, and $W\subseteq X\times Y$, so $\bigcup\mathcal{S}\in\mathcal{P}(X\times Y)$. Let $x\in X$ and $y\in Y$ with $(x,y)\in\bigcup\mathcal{S}$, and take $W\in\mathcal{S}$ with $(x,y)\in W$. Since $W\in\mathcal{T}_{X\times Y}$, there are $U\in\mathcal{T}_{X}$ and $V\in\mathcal{T}_{Y}$ with $x\in U$, $y\in V$ and $U\times V\subseteq W\subseteq\bigcup\mathcal{S}$. Hence $\bigcup\mathcal{S}\in\mathcal{T}_{X\times Y}$.

Intersections. Let $W_{1},W_{2}\in\mathcal{T}_{X\times Y}$; then $W_{1}\cap W_{2}\in\mathcal{P}(X\times Y)$. Let $x\in X$ and $y\in Y$ with $(x,y)\in W_{1}\cap W_{2}$. Take $U_{1}\in\mathcal{T}_{X}$ and $V_{1}\in\mathcal{T}_{Y}$ with $x\in U_{1}$, $y\in V_{1}$ and $U_{1}\times V_{1}\subseteq W_{1}$, and then $U_{2}\in\mathcal{T}_{X}$ and $V_{2}\in\mathcal{T}_{Y}$ with $x\in U_{2}$, $y\in V_{2}$ and $U_{2}\times V_{2}\subseteq W_{2}$. Put $U=U_{1}\cap U_{2}$ and $V=V_{1}\cap V_{2}$. Then $U\in\mathcal{T}_{X}$ and $V\in\mathcal{T}_{Y}$ by Topological Space §intersections, $x\in U$ and $y\in V$. An element of $U\times V$ is a pair $(u,v)$ with $u\in U$ and $v\in V$; for each $i\in\{1,2\}$, $u\in U_{i}$ and $v\in V_{i}$, so $(u,v)\in U_{i}\times V_{i}\subseteq W_{i}$. Thus $U\times V\subseteq W_{1}\cap W_{2}$, and $W_{1}\cap W_{2}\in\mathcal{T}_{X\times Y}$.

The three conditions hold, so $\mathcal{T}_{X\times Y}$ is a topology on $X\times Y$ by Topological Space §topology.