A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis

Assume (B1) and (B2).

Assume instead that $\mathcal{B}$ is a basis for some topology $\mathcal{T}$ on $X$.

$X\in {\mathcal{B}}_{\mathcal{S}}$, being the value of the empty intersection, and $\mathcal{S}\subseteq {\mathcal{B}}_{\mathcal{S}}$, each $S\in \mathcal{S}$ being the intersection of the one-term list $S$.

${\mathcal{B}}_{\mathcal{S}}$ is closed under binary intersections: the intersection of $S_1\cap \cdots \cap S_n$ with $S_1^{\mathrm{\prime }}\cap \cdots \cap S_m^{\mathrm{\prime }}$ is the intersection of the concatenated list, again a list of finitely many members of $\mathcal{S}$.

$\mathcal{B}\subseteq {\mathcal{T}}_{\mathcal{B}}$ always, since for $B\in \mathcal{B}$ and $x\in B$ the set $B$ itself witnesses the defining condition; and every $U\in {\mathcal{T}}_{\mathcal{B}}$ equals $\bigcup \{B\in \mathcal{B}:B\subseteq U\}$, since each such $B$ lies in $U$ and each $x\in U$ lies in one of them.

Under the assumption of step 1.1: $\mathrm{\varnothing }\in {\mathcal{T}}_{\mathcal{B}}$, the defining condition being vacuous, and $X\in {\mathcal{T}}_{\mathcal{B}}$, since by (B1) every $x\in X$ lies in some $B\in \mathcal{B}$ and every subset of $X$ satisfies $B\subseteq X$; so (T1) holds for ${\mathcal{T}}_{\mathcal{B}}$.

Under the assumption of step 1.1: if ${\mathcal{S}}^{\mathrm{\prime }}\subseteq {\mathcal{T}}_{\mathcal{B}}$ and $x\in \bigcup {\mathcal{S}}^{\mathrm{\prime }}$, then $x\in U$ for some $U\in {\mathcal{S}}^{\mathrm{\prime }}$, and membership of $U$ supplies $B\in \mathcal{B}$ with $x\in B\subseteq U\subseteq \bigcup {\mathcal{S}}^{\mathrm{\prime }}$; so $\bigcup {\mathcal{S}}^{\mathrm{\prime }}\in {\mathcal{T}}_{\mathcal{B}}$ and (T2) holds.

Under the assumption of step 1.1: if $U,V\in {\mathcal{T}}_{\mathcal{B}}$ and $x\in U\cap V$, fix $B_1,B_2\in \mathcal{B}$ with $x\in B_1\subseteq U$ and $x\in B_2\subseteq V$; then $x\in B_1\cap B_2$, and (B2) supplies $B_3\in \mathcal{B}$ with $x\in B_3\subseteq B_1\cap B_2\subseteq U\cap V$, so $U\cap V\in {\mathcal{T}}_{\mathcal{B}}$ and (T3) holds.

Under the assumption of step 1.2: $X\in \mathcal{T}$ by (T1), so [L2] gives for each $x\in X$ a member $B\in \mathcal{B}$ with $x\in B\subseteq X$, whence $\bigcup \mathcal{B}=X$, which is (B1); and for $B_1,B_2\in \mathcal{B}\subseteq \mathcal{T}$ the set $B_1\cap B_2$ is open by (T3), so [L2] gives for each $x\in B_1\cap B_2$ a member $B_3\in \mathcal{B}$ with $x\in B_3\subseteq B_1\cap B_2$, which is (B2).

Under the assumption of step 1.2: $\mathcal{T}={\mathcal{T}}_{\mathcal{B}}$. Indeed $U\in \mathcal{T}$ implies $U\in {\mathcal{T}}_{\mathcal{B}}$ by the second form of [L2]; and conversely $U\in {\mathcal{T}}_{\mathcal{B}}$ makes $U$ a union of members of $\mathcal{B}\subseteq \mathcal{T}$ by step 1.5, hence open by (T2).

By steps 1.3 and 1.4, ${\mathcal{B}}_{\mathcal{S}}$ satisfies (B1), since $X\in {\mathcal{B}}_{\mathcal{S}}$ forces $\bigcup {\mathcal{B}}_{\mathcal{S}}=X$, and (B2), since $B_1\cap B_2\in {\mathcal{B}}_{\mathcal{S}}$ may be taken as $B_3$.

Steps 2.1, 2.2 and 2.3 make ${\mathcal{T}}_{\mathcal{B}}$ a topology on $X$ whenever (B1) and (B2) hold, and step 1.5 then makes $\mathcal{B}$ a basis for it and identifies ${\mathcal{T}}_{\mathcal{B}}$ with the family of unions of subfamilies of $\mathcal{B}$; so (B1) and (B2) are sufficient.

Step 2.4 shows (B1) and (B2) are necessary, and step 2.5 shows that any topology having $\mathcal{B}$ as a basis equals ${\mathcal{T}}_{\mathcal{B}}$, which is the asserted uniqueness; with step 3.1 this proves claim 1.

By step 2.6 and step 3.1 applied to ${\mathcal{B}}_{\mathcal{S}}$, the family ${\mathcal{T}}_{{\mathcal{B}}_{\mathcal{S}}}$ is a topology on $X$ with basis ${\mathcal{B}}_{\mathcal{S}}$, and it contains $\mathcal{S}$ by step 1.3 and step 1.5.

Let ${\mathcal{T}}^{\mathrm{\prime }}$ be any topology on $X$ with $\mathcal{S}\subseteq {\mathcal{T}}^{\mathrm{\prime }}$; then ${\mathcal{B}}_{\mathcal{S}}\subseteq {\mathcal{T}}^{\mathrm{\prime }}$, because $X\in {\mathcal{T}}^{\mathrm{\prime }}$ by (T1) covers the empty intersection and (T3) iterated covers the intersections of $n\ge 1$ members of $\mathcal{S}$, and hence ${\mathcal{T}}_{{\mathcal{B}}_{\mathcal{S}}}\subseteq {\mathcal{T}}^{\mathrm{\prime }}$ by (T2), every member of the former being a union of members of ${\mathcal{B}}_{\mathcal{S}}$.

By steps 4.2 and 5.1 the topology ${\mathcal{T}}_{{\mathcal{B}}_{\mathcal{S}}}$ contains $\mathcal{S}$ and is contained in every topology containing $\mathcal{S}$, so it is the coarsest such topology, that is ${\mathcal{T}}_{{\mathcal{B}}_{\mathcal{S}}}=⟨\mathcal{S}⟩$; with step 2.6 this proves claim 2.