Proof
(related to Proposition: Finite Chains are Wellordered)
 Suppose, $(V,\preceq )$ is a finite chain and let $S\subseteq V$ be a nonempty, finite subset of $V$.
 By definition of a chain, "$\preceq$" is a total order, which means that it is, in particular, connex.
 If $x_0\in S$ is the only element of $S$, then $x_0$ is its minimum and we are done. Therefore, suppose $S$ has at least two elements.
 Since "$\preceq$" is connex, for any given $x\in S$ with $x\neq x_0$ we have $x_0\preceq x$ or $x\preceq x_0.$
 Therefore, either $x_0$ is a minimum of $S$ or there is a smaller element $x_1\in S$ with $x_1\preceq x_0.$
 By the same argument, either $x_1$ is a minimum of $S$ or there is a smaller element $x_2\in S$ with $x_2\preceq x_1.$
 Since $S$ is finite, this chain of arguments has to terminate at some $x_i,$ $i\ge 0$ which is the minimum of $S.$
 Therefore, $V$ is wellordered, by definition.
∎
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References
Bibliography
 Reinhardt F., Soeder H.: "dtvAtlas zur Mathematik", Deutsche Taschenbuch Verlag, 1994, 10th Edition