A finite set always has a maximum and a minimum.
Let $S = \{s_1, \ldots,s_n\}$ be a nonempty finite set of size $n > 0$. We will show by induction on $n \in \mathbb N$ that there exist some $m,M \in S$ such that for all $s \in S$, we have that $m \leq s \leq M$.
Base Case: For $n=1$, we have $S = \{s_1\}$, so taking $m = s_1$ and $M=s_1$ trivially satisfies the required condition.
Induction Hypothesis: Assume that the claim holds for $n=k$, where $k \geq 1$.
It remains to prove that the claim holds true for $n = k+1$. To this end, choose any set $S$ with $k+1$ elements, say $S = \{s_1 ,\ldots,s_k,s_{k+1}\}$. Now by the induction hypothesis, the subset: $$ S' = S \setminus \{s_{k+1}\} = \{s_1 ,\ldots,s_k\} $$ has a minimum element and a maximum element. That is, we know that there exists some $m',M' \in S'$ such that for all $s' \in S'$, we have that $m' \leq s' \leq M'$. Now observe that $s_{k+1}$ must fall under $1$ of $3$ cases:
Case 1: Suppose that $s_{k+1} < m'$. Then take $m = s_{k+1}$ and $M=M'$. To see why this works, observe that any element in $S$ is either $s_{k+1}$ or some $s' \in S'$, and: $$ m = s_{k+1} < m' \leq s' \leq M' =M $$
Case 2: Suppose that $m' \leq s_{k+1} \leq M'$. Then take $m = m'$ and $M=M'$. To see why this works, observe that any element in $S$ is either $s_{k+1}$ or some $s' \in S'$, and: $$ m = m' \leq s_{k+1} \leq M' = M $$ $$ m = m' \leq s' \leq M' = M $$
Case 3: Suppose that $s_{k+1} > M'$. Then take $m =m'$ and $M=s_{k+1}$. To see why this works, observe that any element in $S$ is either $s_{k+1}$ or some $s' \in S'$, and: $$ m = m' \leq s' \leq M' < s_{k+1} = M $$
Hence, we have shown that $S$ has a minimum and maximum element, as desired.
Let $F$ be a finite set. if $F$ is $\{x\} $ then we are done since we vacouly have $x \geq x $ and hence $x = \max \{ x \} $. If $F = \{ a_1 ,... a_n \} $. assume they are different, otherwise we are back again to singleton case. Now, take $a_1$. IF $a_1 $ is greater than any other $a_i$ then set $a_1 = \max F $ and we are done. IF not, take $a_2$, and repeat previous step. Continue in this manner inductively. Eventually, we get the max.