The way I understand minimal and maximal is the following.
Consider the following collection of sets:
{{1,2},{2,4},{1,6,7},{1,2,4,5,6},{8}}
The minimal members of this collection of sets is: {1,2},{2,4},{1,6,7},{8}
Meaning there are no other sets in the collection that are proper subsets of the set.
Minimal Set Example 1: in the set {1,2} this case holds true because the only proper subsets possible are {1} and {2}. Those sets {{1},{2}) are nowhere to be found in the collection. Therefore we include {1,2} as a part of the minimal collection of sets.
Minimal Set Example 2: in the set {1,6,7} this case holds true because the only proper subsets possible are {1},{6},{7},{1,6},{1,7}, and {6,7}. Those proper subsets ({1},{6},{7},{1,6},{1,7}, and {6,7}) are nowhere to be found in the collection. Therefore we include {1,6,7} as a part of the minimal collection of sets.
The maximal members of this collection of sets is: {1,6,7},{1,2,4,5,6},{8}
Meaning none of these sets are proper subsets of other sets in the collection. NOTICE: The maximal collection is not mutually exclusive of the minimum collection. See {1,6,7}.
Maximal Set Example 1: in the set {8} this case holds true because this set is not found to be a subset of any of the other sets in the collection. Meaning 8 is not found in any of the other sets in the collection. Therefore we include {8} as a part of the maximal collection of sets.
Maximal Set Example 2: in the set {1,6,7} this case holds true because this set is not found to be a subset of any of the other sets in the collection. Meaning all of the numbers of the set {1,6,7} are not found together as a part of another set in the collection. Therefore we include {1,6,7} as a part of the maximal collection of sets.
Best Answer
For a single-element set, the element is maximal and minimal.
Assume that all $n$-element subsets have a maximal and a minimal element. Let $A$ sub a subset with $n+1$ elements and $x$ one of its elements. Then $B=A\setminus\{x\}$ has $n$ elements. Let $M,m$ be maximal and minimal elements of $B$. Then $\max(x,M)$ is maximal and $\min(x,m)$ is minimal. In fact, if $y\in A$ and $y\geq \max(M,x)$, then either $y\in B$ or $y=x$. In the first case, it follows that $y=M$ and $M\geq x$. Therefore, $y=\max(M,x)$. In the second case, it follows that $y=x\geq M$ and therefore $y=\max(M,x)$. The argument for $\min(m,x)$ is similar but with the inequalities reversed.
Note that the argument actually shows the existence of maximum and minimum, since the set is assumed to be totally ordered.
Note that although the question includes the condition of totally ordered, the argument can be adapted to also work in this case. What is needed is to replace $\max(M,x)$ and $\min(m,x)$ for $M$ and $n$, in the case that $M$ and $x$ are not comparable and the case that $m$ and $x$ are not comparable, respectively. In this case, one only gets the existence of maximal and minimal elements, but not necessarily maximum and minimum.