I am kind of stuck on how to prove this. I thought maybe cases would work?
Let $S$ be a nonempty subset of a ring $R$. Show that there exists a subring $R'$ of $R$ that contains $S$.
Case 1: $S$ is a ring, so $R'=S$.
Case 2: $S$ is not a ring, so construct $R'$ to contain $S$ and elements needed to be a subring of $R$.
This is what my only idea is so far. Is there any easier or better way to prove this?
$R$ is a subring of itself so you can always let $R' = R$. If you mean for $R'$ to be a proper subring of $R$ then the claim is not true. For a counterexample, let $S = \mathbb{Z} - \{0\}$ in the ring $\mathbb{Z}$. Then the only subring of $\mathbb{Z}$ containing $S$ is $\mathbb{Z}$ itself.