Proof of $\sup(1,3) \cup (5, 8) = 8$. Proof:
Let $S = (1,3)\cup(5,8)$
Clearly $\forall x \in S,\;x<8$. Hence $8$ is an upper bound of $S$.
To prove that 8 is the least upper bound suppose there exists an upper bound less than $8$.
So $\exists m<8 \;s.t \;\forall x \in S,\;x<m$.
But $m<\frac{m+8}2 < 8$ and $\frac{m+8}2 \in S$.
Therefore $m$ is not an upperbound. Hence $8$ is the supremum of $S$.  
Is the proof correct ?
I have seen in some proofs, they choose an upper bound $m$ as above and take $x = m+\frac{8-m}2$. Then prove $m < x$  and $x \in S$. Is my proof simpler than that ?
 A: Your proof is very similar to the other proof; note that
$$m + \frac{8 - m}{2} = \frac{2m}{2} + \frac{8 - m}{2} = \frac{8 + m}{2},$$
i.e. you're using the same counter-example as the other proof, just written differently.
As for whether this proof works, I would say it doesn't quite work. If I gave you the possible upper bound $m = 0$ (I know it isn't an upper bound, but still) and naively plugged it into your argument, then $\frac{m + 8}{2} = 4 \notin S$, contrary to your claim.
You might want to modify the expression $m + \frac{8 - m}{2}$ to $m + ???$ where $???$ is some expression involving $\frac{8 - m}{2}$ and a max formula...
A: As was noted, the main problem is whether the argument for $\frac{m+8}{2}\in S$ is valid. We suppose $m<8$ is an upper bound of $S$. We have to show $1<\frac{m+8}{2} < 3$ or $5<\frac{m+8}{2}<8$.  
Suppose it's not the first one. As $m<8$ we have $\frac{m+8}{2}<\frac{8+8}{2} = 8$. If $\frac{m+8}{2} \leq 1$, then $m\leq -6$, which is impossible, because $m$ is an upper bound of $S$, so we must have $\frac{m+8}{2} \geq 3$. If $\frac{m+8}{2} \leq 5$, then $m\leq 2$, which again would be impossible. Thus $\frac{m+8}{2} >5$.
We also see that $m = \frac{m+m}{2} < \frac{m+8}{2} < 8$, which contradicts $m$ being an upper bound of $S$.

Note that it doesn't matter what $m$ is, exactly. All that matters is the supposition $m<8$ enables us to define a certain element in $S$, which gives us a contradiction.
