Note that the intersection of any nonempty family of transitive relations on $A$ is itself a transitive relation on $A$: indeed, if $S_i$, $i\in I$, is a nonempty family of transitive relations on $A$, and $T=\cap_{i\in I}S_i$, then $T$ is a relation on $A$; and if $(a,b),(b,c)\in T$, then for every $i\in I$ we have $(a,b),(b,c)\in S_i$. Since each $S_i$ is transitive, then $(a,c)\in S_i$ for each $i$, hence $(a,c)\in \cap_{i\in I}S_i = T$. Thus, $T$ is a transitive relation.
That means that you can let $T = \bigcap \bigl\{ R\subseteq A\times A \bigm| \cup_{S\in U}S\subseteq R\text{ and }R\text{ is transitive}\bigr\}.$ Since $A\times A$ is an element of this set, the intersection is well-defined, and $T$ is the smallest relation on $A$ that is transitive and contains each $S_i$.
The relation $T$ is called the "transitive closure of $\cup_{S\in U}S$". The cardinality of $U$ is irrelevant.
That's the "top-down" construction. The "bottoms up" construction is that you need to take all pairs $(a,b)$ such that there exists a positive integer $n$, and elements $x_1,\ldots,x_n\in A$ (not necessarily distinct), and relations $S_{1},\ldots,S_n$ in $U$ (not necessarily distinct) such that $(a,x_1)\in S_1$, $(x_1,x_2)\in S_2,\ldots, (x_{n-1},x_n)\in S_n$, and $x_n=b$. Then it is an easy exercise to show that this set contains each $S_i$ and is transitive, and that any transitive relation on $A$ that contains each $S_i$ must contain this set, so this set is the transitive closure of the $S_i$. Note that the construction only uses finitely many elements of $U$ at a time, so the cardinality of $U$ is irrelevant: it can be as large as you want.