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 $S_1 \cap \dots \cap S_n$ with $S'_1 \cap \dots \cap S'_m$ 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: $\varnothing \in \mathcal{T}_{\mathcal{B}}$, 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 $\mathcal{S}' \subseteq \mathcal{T}_{\mathcal{B}}$ and $x \in \bigcup \mathcal{S}'$, then $x \in U$ for some $U \in \mathcal{S}'$, and membership of $U$ supplies $B \in \mathcal{B}$ with $x \in B \subseteq U \subseteq \bigcup \mathcal{S}'$; so $\bigcup \mathcal{S}' \in \mathcal{T}_{\mathcal{B}}$ 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 $\mathcal{T}'$ be any topology on $X$ with $\mathcal{S} \subseteq \mathcal{T}'$; then $\mathcal{B}_{\mathcal{S}} \subseteq \mathcal{T}'$, because $X \in \mathcal{T}'$ by (T1) covers the empty intersection and (T3) iterated covers the intersections of $n \ge 1$ members of $\mathcal{S}$, and hence $\mathcal{T}_{\mathcal{B}_{\mathcal{S}}} \subseteq \mathcal{T}'$ 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 $\mathcal{T}_{\mathcal{B}_{\mathcal{S}}} = \langle \mathcal{S} \rangle$; with step 2.6 this proves claim 2.