Proof of The Subspace Topology is a Topology

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. By its definition $\mathcal{T}_{A}$ is a subset of $\mathcal{P}(A)$, and a set by Sets and Maps: Ordinary Notation §set-builder, as Topological Space §topology requires; we check the conditions Topological Space §empty-whole, Topological Space §unions and Topological Space §intersections for $\mathcal{T}_{A}$ on $A$.

Empty set and whole set. By Topological Space §empty-whole for $\mathcal{T}$, $\emptyset\in\mathcal{T}$ and $X\in\mathcal{T}$. The subsets $\emptyset$ and $A$ of $A$ satisfy $\emptyset=A\cap\emptyset$ and, as $A\subseteq X$, $A=A\cap X$; so $\emptyset\in\mathcal{T}_{A}$ and $A\in\mathcal{T}_{A}$.

Unions. Let $\mathcal{S}$ be a subset of $\mathcal{T}_{A}$, and let $\mathcal{S}'=\{U\in\mathcal{T}:A\cap U\in\mathcal{S}\}$, a set by Sets and Maps: Ordinary Notation §set-builder. As $\mathcal{S}'\subseteq\mathcal{T}$, Topological Space §unions gives $\bigcup\mathcal{S}'\in\mathcal{T}$. We show $\bigcup\mathcal{S}=A\cap\bigcup\mathcal{S}'$. If $u\in A\cap\bigcup\mathcal{S}'$, then $u\in U$ for some $U\in\mathcal{S}'$, so $u\in A\cap U$ with $A\cap U\in\mathcal{S}$, and $u\in\bigcup\mathcal{S}$. Conversely, if $u\in\bigcup\mathcal{S}$, then $u\in W$ for some $W\in\mathcal{S}$; since $W\in\mathcal{T}_{A}$ there is $U\in\mathcal{T}$ with $W=A\cap U$, so $U\in\mathcal{S}'$ and $u\in A\cap U\subseteq A\cap\bigcup\mathcal{S}'$. Hence $\bigcup\mathcal{S}=A\cap\bigcup\mathcal{S}'$, a subset of $A$ of the required form, and $\bigcup\mathcal{S}\in\mathcal{T}_{A}$.

Intersections. Let $W_{1},W_{2}\in\mathcal{T}_{A}$, and take $U_{1},U_{2}\in\mathcal{T}$ with $W_{1}=A\cap U_{1}$ and $W_{2}=A\cap U_{2}$. Then $U_{1}\cap U_{2}\in\mathcal{T}$ by Topological Space §intersections, and $W_{1}\cap W_{2}=A\cap(U_{1}\cap U_{2})$, a subset of $A$; so $W_{1}\cap W_{2}\in\mathcal{T}_{A}$.

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