Suppose that a set $E \subset S$ is not connected, then it can be written as the union of two separated sets $A$ and $B$. By separated, I mean $A \cap \bar{B} = \bar{A} \cap B = \emptyset$.
My question is are $\bar{A}$ and $\bar{B}$ viewed as closures of $A$ and $B$ in the induced topology $E$ or are they viewed as closures in the full set $S$, or are they equivalent, or does it matter?
It's equivalent. We can view $E$ as a topological space by itself and then wonder if this space is connected or not. If it is disconnected, it can be written as disjoint union of two nonempty open sets. Back in the langugage of relative topology this means that we can find two open sets $U,V\subseteq S$ such that $U\cap E$ and $V\cap E$ are disjoint. It is however a misconception to assume that $U\cap V=\emptyset$. With your formulation, this pitfall is avoided: The closure of $B$ in $E$ is the closure of $B$ in $S$, intersected with $E$; hence as $A$ is a subset of $E$, whether $A\cap \bar B=\emptyset$ does not depend on whether we consider closure in $E$ or in $S$. However, we may conclde $\bar A\cap\bar B=\emptyset$ only in the suspace topology, not in $S$.
