Proposition: A nonempty subset $S$ of $\mathbb{N}$ with a single element $\eta$ is bounded above and below.
Proof. Suppose $S$ does not have an upper bound. Then, there exists no number $u$ s.t. $u \gt \eta$. But $\eta + 1 > \eta$, which is a contradiction. Therefore, $S$ is bounded above.
Suppose $S$ does not have a lower bound. Then, there exists no number $\ell$ s.t. $\eta > \ell$. But $\eta > \eta - 1$, which is a contradiction. Therefore, $S$ is bounded below. $\square$
Would this be satisfactory or am I missing something? I'm attempting to learn how to write acceptable proofs in real analysis.
It's OK, but there is no need to use contradiction. You can just say $\eta$ is both an upper bound and a lower bound (because $\eta \le \eta$ and $\eta \ge \eta$).