how to prove finite set of natural number A has a largest number by contradiction?

I think you do not to proove $M=\mathbb N$. Because if $A\ne\emptyset$, then there exists $a$ as an element of $A$, and $a\notin M$.

I'll use another variation of $M$ you used, as $L=\{n\in\mathbb N\mid\forall a\in A, a\le n\}$. Then $L$ is not an empty set, because $A$ is finite, so $\sum_{a\in A}a=x\in L$.

And by well-orderness of $(\mathbb N, \le)$, every non-empty subset of $\mathbb N$ has their minimal element. I'll let $\min L=m$.

Then $\forall a\in A, a\le m$. And if $m\notin A$, then $\forall a\in A, a< m$ holds. This mean $\forall a\in A, a\le m-1$. So $m-1\in L$. This contradicts with minimality of $m$. So $m\in A$.

So, $m$ is the maximal element of $A$(what means, $m$ is the largest number of $A$).