My textbook defines a theory as a set of formulas that is closed under logical consequence. According to the book, a set of formulas U is closed under logical consequence iff for all formulas A, if $$U \vDash A $$then, $$A \in U$$ Does this mean that a set of formulas is closed under logical consequence if there exists a satisfying interpretation for U?
Would this set be closed under logical consequence? $$ U = \{\neg p \vee q, p \wedge (q \vee \neg r), r\} $$
No, your $U$ is not closed under logical consequence. For example, the statement $p$ is a logical consequence of $U$, but it is not in $U$. Same for $p \land p$, $p \land p \land p$, $z \lor \neg z$, etc.
Indeed, any set closed under logical consequence has to be of infinite size, so given that this set is of finite size tells you immediately that it is not closed under logical consequence.
Finally: you are trying to make some connection between satisfiability and logical consequence. I am not sure what connection you see, but there is this: if a set is not satisfiable, then every statement is a logical consequence of that set, and hence every statement should be in the logical closure of that set. In other words, a set is not satisfiable and closed under logical consequence if and only if it is the set of all logic statements.