This is what I need to prove:
Let $S$ be a finite set of reals. If $n>s$ for every $s\in S$ then $n>\max S$.
Here's a proof by contradiction that I tried:
Let $s\in S$. Then $n>s$. By our assumption, $n\le \max S$. Hence we have $s<\max S$. Since $s$ was arbitrarily chosen, it must be the case that $\max S > s$ for every $s \in S$. Thus $\max S \not\in S$. This leads to a contradiction.
Is this proof correct? Are there alternative ways to prove this?
It's fine, but there is no need to appeal to contradiction. $\mathrm{ max} S \in S$ by definition, and so $n>\mathrm{max} S$ by assumption.