Your T is reminiscent to me of constructing a *sub*basis (rather than a basis) for a topology S.
[The next few definitions may already be familiar to you. If so, I include them for the benefit of other readers.] Let X be a set, and let S be a subset of the powerset of X containing at least the empty set and X itself. (For convenience, we define an intersection with zero operands to be X, which conforms with the usual rule that an associative operator applied to zero operands equals the identity for that operator.)
1. Definition of topology: Suppose that S contains (as elements) 1.1) all pairwise intersections of its elements and 1.2) all unions (not necessarily pairwise) of its subsets. Then S is called a topology on X and the elements of S are called the open sets. Note that 1.1 implies that S contains all intersections of its finite subsets.
2. Definition of basis (base is also commonly used) of a topology: Let S be a topology on X. Let B be a subset of S such every element of S (i.e., every open set of X) is a union of a subset of B. Then B is called a basis for S.
3. Definition of subbasis (or subbase) of a topology: Let S be a topology on X and let C be a subset of S such that every element of S is a union (not necessarily of a finite number of opetands) of intersections of finite subsets of C. Then C is called a subbasis for S.
Quick examples (that should be familiar to almost all readers): Let R be the real numbers. Let's declare a subset A of R to be open if for all x in A there is a d > 0 such that the open interval (x-d, x+d) is a subset of A. Then the set of all such A is a topology on R (indeed it's the standard topology). The set of all open intervals is a basis for that topology. The set of all open intervals that are unbounded at one end is a subbasis for that topology. (These things require some proof, which is essy, the subbasis part being trivial once the topology and basis are verified).
A place where this analogy between topology and your S and T breaks down is that the elements of a topological subbasis are not necessarily pairwise disjoint. However, it may be worthwhile to restrict your attention to cases where the union of S (in other words, X) is finite.