Let S be an ordered set. Let $A \subset S$ be a nonempty finite subset. Prove that $A$ is bounded, that inf A exists and is in $A$, and that $\sup (A)$ exists and is in $A$. Hint: Use induction. (A finite set is a set from which, for some $n \in \mathbb{N}$, there is an injective map to the set {$1, 2, \cdots , n$}
I started with induction on cardinality on set $|A|$. But I am stuck right here in the beginning.
If $A$ has one member, it's obvious that it's bounded, and
$$\text{inf } A = \text{sup } A \in A.$$
Assuming the statements hold for an $n$-member subset, add another member.
The new member exists, since it's a member of $S$.
Since $A$ is bounded, the addition of one member does not change this (a finite set plus one member is still a finite set). (Call the new set $A'$, and the new element $m$.)
This new member $m$ is either greater than $\text{sup } A$ (which exists), or it is less than $\text{sup } A$. If it is greater, $\text{sup } A' = m$. Otherwise, $\text{sup } A' = \text{sup } A.$
Also, this new member $m$ is either less than $\text{inf } A$ (which exists), or it is greater than $\text{inf } A$. If it is less, $\text{inf } A' = m$. Otherwise, $\text{inf } A' = \text{inf } A.$
I think that covers everything?