Proof Verification: Tao, Analysis I: Exercise 5.5.3. Let $E$ be a subset of $\Bbb{R}$, let $n\geq 1$ be an integer, and let $m,m'\in\Bbb{Z}$ with properties that $\frac{m}{n}$ and $\frac{m'}{n}$ are upper bounds of $E$, but $\frac{m-1}{n}$ and $\frac{m'-1}{n}$ are not upper bounds. Then $m=m'$.
Proof:(by contradiction)
Suppose $E\subset\Bbb{R}$, $m,m'\in\Bbb{Z}$, and $n\in\Bbb{Z^+}$ such that $\forall \xi\in E$, $\xi\leq \frac{m}{n},\frac{m'}{n}$ and $\exists\overline{\xi}\in E$ such that $\frac{m-1}{n},\frac{m'-1}{n}\leq \overline{\xi}$. If $m\neq m'$ then either $m<m'$ or $m>m'$. Without loss of generality, let $m'<m$. Since $\forall n\in\Bbb{Z^+}$, $\frac{1}{n}\in\Bbb{Q^+}$ then 
\begin{align}\frac{m'}{n}<\frac{m}{n} &\Longleftrightarrow\frac{m'-1}{n}<\frac{m-1}{n}<\frac{m'}{n}<\frac{m}{n}\\ &\Longleftrightarrow \exists\overline{\xi}\in E(\frac{m'-1}{n}<\frac{m-1}{n}<\overline{\xi}<\frac{m'}{n}<\frac{m}{n})\end{align}
We want to show that if $m'<m$ then $\forall\xi\exists\overline{n}\in\Bbb{Z^+}(\frac{m'-1}{n}<\overline{\xi}<\frac{m-1}{n})$ to yield a contradiction. Considering,
\begin{align}
\frac{m'-1}{n}<\overline{\xi}<\frac{m-1}{n} &\Longleftrightarrow \frac{m'-1}{n}<\overline{\xi}\quad and\quad \frac{m-1}{n}<\overline{\xi}\\ &\Longleftrightarrow \frac{m'-1}{\overline{\xi}}<\overline{n}<\frac{m-1}{\overline{\xi}}
\end{align}
If $\vert \frac{m-1}{n}-\frac{m'-1}{n}\vert >1$, then $\exists\overline{n}\in\Bbb{Z}$ that satisfies the above conditions. This contradicticts the premise that for all $\overline{\xi}\in E$ such that $n\geq 1$ then $\frac{m-1}{n}$ and $\frac{m'-1}{n}$ are not upper bounds.

Your proof seems to be o.k. But the following arguments are simpler:
Suppose that $m'<m$. Then $m-m'>0$. Since $m-m' \in \mathbb Z$, we have $m-m' \ge 1.$  Thus $m' \le m-1$ and therefore $\frac{m'}{n} \le \frac{m-1}{n}.$
Conclusion: $\frac{m-1}{n}$ is an upper bound of $E$. Contradiction !
Hence we have $m' \ge m$. The inequality $m \ge m'$ can be proved in a similar way.
