Proof of The Subspace Topology is a Topology

Every element of $\mathcal{T}_A$ has the form $A\cap U$ with $U\in\mathcal{T}$ and is therefore a subset of $A$, so $\mathcal{T}_A$ is a collection of subsets of $A$. We verify the three conditions of Topological Space for the pair $(A,\mathcal{T}_A)$.

Condition 1. By condition 1 of Topological Space applied to $(X,\mathcal{T})$ one has $\emptyset\in\mathcal{T}$ and $X\in\mathcal{T}$. Since $\emptyset=A\cap\emptyset$ and $A=A\cap X$, both $\emptyset$ and $A$ belong to $\mathcal{T}_A$.

Condition 2. Let $B$ be a set and let $(V_b)_{b\in B}$ be a family of subsets of $A$ with $V_b\in\mathcal{T}_A$ for every $b\in B$. Put

and for $p=(b,U)\in P$ set $W_p=U$. Then $(W_p)_{p\in P}$ is a family of subsets of $X$ with $W_p\in\mathcal{T}$ for every $p\in P$, so by condition 2 of Topological Space the set

belongs to $\mathcal{T}$. Note that no choice is used here: $P$ collects all admissible pairs rather than one witness per index. Since $V_b\in\mathcal{T}_A$, for every $b\in B$ there is at least one $U\in\mathcal{T}$ with $V_b=A\cap U$, so every $b\in B$ occurs as the first coordinate of some element of $P$.

We claim that

If $x\in A\cap W$, then $x\in W_p$ for some $p=(b,U)\in P$, and $W_p=U$, so $x\in A\cap U=V_b$. Conversely, let $x\in V_b$ for some $b\in B$, and let $(b,U)\in P$ be an element with first coordinate $b$. Then $x\in V_b=A\cap U$, and $U=W_{(b,U)}\subseteq W$, so $x\in A\cap W$. This proves the claim, and hence $\bigcup_{b\in B}V_b=A\cap W\in\mathcal{T}_A$.

Condition 3. Let $n$ be a natural number and let $V_1,\dots,V_n\in\mathcal{T}_A$. Let $[n]$ denote the initial segment determined by $n$, and for $k\in[n]$ put

Then $(C_k)_{k\in[n]}$ is a family of subsets of $\mathcal{T}$, and each $C_k$ is nonempty because $V_k\in\mathcal{T}_A$. By claim 1 of Basic Properties of Finite Sets the set $[n]$ has $n$ elements, so it is finite. Hence Choice for a Family Indexed by a Finite Set yields a function $c:[n]\to\mathcal{T}$ with $c(k)\in C_k$ for every $k\in[n]$; writing $U_k=c(k)$ we obtain $U_1,\dots,U_n\in\mathcal{T}$ with $V_k=A\cap U_k$ for every $k\in[n]$.

By condition 3 of Topological Space the set

belongs to $\mathcal{T}$. For $x\in X$ we have $x\in A\cap U$ if and only if $x\in A$ and $x\in U_k$ for every $k\in[n]$, which holds if and only if $x\in A\cap U_k=V_k$ for every $k\in[n]$. Therefore

and consequently $\bigcap_{k=1}^{n}V_k\in\mathcal{T}_A$.

All three conditions hold, so $(A,\mathcal{T}_A)$ is a topological space.