 Solution

Well-orders are not first-order axiomatisable. Suppose a theory $T$ axiomatized them. Enlarge the language by constants $c_0,c_1,\dots$ and addEvery finite subset of the resulting theory has a model in the well-order $(\mathbb{N},<)$: assign its finitely many constrained constants to a sufficiently long finite descending chain. Thus the compactness theorem supplies a model of $T$ with an infinite descending chain. The set of the $c_i$ has no least element, contradicting well-ordering. This shows why the second-order assertion that every nonempty subset has a least member cannot be replaced by a first-order theory in the poset language.