= Solution
The class of well-orderings is \b[not first-order axiomatizable]. If $S$ axiomatized it, add constants $c_0,c_1,\ldots$ and sentences
$$
c_{n+1}<c_n\qquad(n\geq0).
$$
Every finite subset has a finite well-order containing the required finite descending chain. By the <compactness theorem>, the whole set has a model; that model satisfies $S$ but contains an infinite descending chain, so it is not a <well-order>. This contradiction proves the claim, as summarized by <well-orders are not first-order axiomatizable>.
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