The Subgroup Generated By S Is Actually A Subgroup

Any time you want to show something is a subgroup, the subgroup test (see problem (Subgroups Are Subsets That Are Closed Under Products And Inverses)) will simplify your work. So you just need to show that $\left<S\right>$ is a nonempty subset of $G$ and if $a,b\in \left<S\right>$, then both $ab\in \left<S\right>$ and $a^{-1}\in \left<S\right>$.