$(A,\leq)$ is a partially ordered set. That each finite non-empty subset of $A$ has a $maxB$ implies $(A,\leq)$ is totally ordered.

Question

My textbook contains the following exercise:

Let $(A,\leq)$ be a partially ordered set. Prove that if each finite non-empty subset of $A$ has a greatest element, then $(A,\leq)$ is totally ordered.

I'm struggling to prove this.
I began by just considering some subset of $A$ containing two elements. I denote this subset of $A$ as $B_1$: $B_1 \subseteq A$. I denote the two elements in $B_1$ as $x_1$ and $x_2$: $B_1$={$x_1, x_2$}. Assuming the antecedent of the above conditional sentence, it's true that $x_1$ or $x_2$ is the $maxB_1$. Then $x_2 \ge x_1$ or $x_1 \ge x_2$. Now, suppose $x_1=x_2$. Then, it's clear that $B_1$ is totally ordered. Then, it remains to show that in the cases $x_1 \lt x_2$ and $x_2 \lt x_1$, $x_1$ and $x_2$ are $\le$-comparable. (Establishing this fact essentially establishes the rest of the proof, well, at least to my thinking.) It is at this point that I'm stuck.

Answer 1

HINT: Turn your argument around. Suppose that $A$ is not totally ordered. Then there are $x_1,x_2\in A$ such that $x_1\not\le x_2$ and $x_2\not\le x_1$. What can you say about the set $\{x_1,x_2\}$?

Answer 2

The only difference between a partial order and a total order is that in a total order any two elements can be compared. But if every set of two elements has a maximum, by definition of maximum it means those two elements can be compared.