Proving supremum through midpoints?

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 ?