I have been trying to formulate a good proof that the supremum of a set is unique. I came up with this proof, however I am not sure if it is correct.
Prof by contradiction: $$ \exists M_0 \neq M_1, M_0 \ and \ M_1 are \ supremums \ of A $$ $$ \forall \epsilon > 0, \exists a \in A : M_0 - \epsilon < a $$ $$ \forall \epsilon > 0, \exists a \in A : M_1 - \epsilon < a $$ $$ \Rightarrow \begin{cases} M_0−ϵ<a \\ M_1−ϵ<a \end{cases} $$ $$ \Rightarrow \begin{cases} M_0−M_1<0 \Leftrightarrow M_0<M_1\\ M_1−M_0<0 \Leftrightarrow M_1<M_0 \end{cases} $$ this is a contradiction, thus $M_{0}$ = $M_{1}$
No, that is not correct. From$$\left\{\begin{array}{l}M_0-\varepsilon<a\\M_1-\varepsilon<a\end{array}\right.\tag1$$you cannot deduce that $M_0-M_1<0$. For instance, if $M_0=-1$, $M_1=-2$, $\varepsilon=1$, and $a=0$, then $(1)$ holds. However, $M_0-M_1=1>0$.
However, you can take $\varepsilon=|M_0-M_1|$. If $M_1>M_0$, take $a\in A$ such that $M_1-\varepsilon<a\leqslant M_1$. But then $M_1-\varepsilon>M_1-(M_1-M_0)=M_0$. Then $a>M_0$, which is impossible, since $a\in A$ and $M_0=\sup A$. The case in which $M_1<M_0$ is similar.