Every finite set contains its supremum: proof improvement.

Your proof starts right away with using $\max A$. But if you know the existence of $\max$ then it is automatically also the $\sup$. So this is probably not the way you are expected to proceed.

If $A=\{a\}$ is a singleton set, then clearly $\sup A=\max A = a$.

If $A$ has cardinality $n>1$ and we know as induction hypothesis that all sets of cardinality $<n$ have a maximal element, let $a\in A$ be an arbitrary element and let $A'=A\setminus\{a\}$. Since $A'$ has less than $n$ elements, let $a'= \max A'$. If $a'\ge a$, then $a'$ is a maximal element of $A$. If $a'< a$, then $a$ is a maximal element of $A$.