A set is a finite chain if every subset has a top and bottom element I am presently attempting Exercise 2 in Kaplansky, Set Theory and Metric Spaces
Exercise 2: Let $L$ be a partially ordered set in which every subset has a top and bottom element. Prove that $L$ is a finite chain.
Proof: Denote a subset of $L$ by $S$. If $S$ has a top and bottom element, then $\sup S$ and $\inf S$ exist and are elements of $S$. Denote them $a$ and $A$ respectively. Since $a,A \in S$ then $S$ is finite. This means that all the elements in $S$ can be ordered from smallest to largest, thus we have: $S= \{ a, ..., A \}$. Based upon this ordering, given any two elements, $b,c \in S$ one may determine that $b \le c$ or $c \le b$. Since all the subsets of L are a chain, this implies that L must also be a chain.
My Question: I am uncertain on whether the step in bold is valid. I believe it is based upon the transitivity condition in the definition for a partially ordered set, though I would appreciate some feedback on the matter.
 A: Given any two elements $x,y\in L,$ we know that $\{x,y\}$ has a top element and a bottom element. This shows that comparability holds on $L$, so since $L$ is a poset, then $L$ is a linearly ordered set.
Now, suppose that $L$ is not finite. Let $x_0$ be the top element of $L$, and for any nonnegative integer $n,$ let $x_{n+1}$ be the top element of $L\setminus\{x_0,...,x_n\}.$ Show that $\{x_n:n\text{ a nonnegative integer}\}$ is a subset of $L$ without a bottom element--the desired contradiction (you'll need the assumption that $L$ isn't finite).
P.S. (Added later): The actual claim should read "...every non-empty subset has...," for obvious reasons. In fact, we can adjust the claim as follows:

Let $L$ be a partially ordered set. Then $L$ is a finite chain if and only if every non-empty subset of $L$ has a top element and a bottom element.

The forward direction can be proved by induction, though this may vary in difficulty, depending on which definition of "finite" is being used.
A: No, it is not valid.  There are partially ordered sets, finite or infinite, that are not chains.  (E.g., consider a power set of a set with at least $2$ elements, ordered by inclusion.)  You have to use the hypothesis to show that $L$ is a chain.  I recommend contraposition.  You can show that if there are $2$ incomparable elements, then there exists a nonempty subset without a top and bottom.
You also gave no valid reason why $L$ is finite.  
