The lexicographic order on $\mathbb{Z}\times \mathbb{Z}$ is not a well-order

The relation $\prec$ is irreflexive: $(a,b)\prec (a,b)$ would require $a<a$, or $a=a$ and $b<b$, and both alternatives fail because the strict order of $\mathbb{Z}$ is irreflexive.

At least one of $\prec$, $=$, $\succ$ links two pairs $(a,b)$ and $(c,d)$: by totality of the order of $\mathbb{Z}$ either $a<c$, or $a=c$, or $c<a$; in the first case $(a,b)\prec (c,d)$, in the last $(c,d)\prec (a,b)$, and in the middle case totality of the order of $\mathbb{Z}$ decides between $b<d$, $b=d$ and $d<b$, giving $(a,b)\prec (c,d)$, $(a,b)=(c,d)$, and $(c,d)\prec (a,b)$ respectively.

The relation $\prec$ is transitive: if $(a,b)\prec (c,d)$ and $(c,d)\prec (e,f)$, then by the definition of $\prec$ either $a<c$, or $a=c$ and $b<d$, and either $c<e$, or $c=e$ and $d<f$. If $a<c$ then $c\le e$ in both second alternatives, so $a<e$ by transitivity of the order of $\mathbb{Z}$; if $a=c$ and $b<d$, then $c<e$ gives $a<e$, while $c=e$ gives $a=e$ together with $b<d<f$, hence $b<f$. In every case $(a,b)\prec (e,f)$.

The set $\{0\}\times \mathbb{Z}$ is a nonempty subset of $\mathbb{Z}\times \mathbb{Z}$, and its elements are exactly the pairs $(0,n)$ with $n\in \mathbb{Z}$; for instance $(0,0)$ belongs to it.

$0<1$ in $\mathbb{Z}$: the integer $1$ is $[(1,0)]$ and $0$ is $[(0,0)]$ by the operations of [F3]; since $1=\sigma (0)\mathrm{\ne }0$ in $\mathbb{N}$, the representatives $[(1,0)]$ and $[(0,0)]$ are distinct, because the canonical embedding $n\mapsto [(n,0)]$ is injective by [F4]; and $[(1,0)]$ lies in the image of that embedding, which is the set of nonnegative integers, so $0\le 1$ in $\mathbb{Z}$; therefore $0<1$ by [F1].

At most one of the three alternatives holds: $(a,b)\prec (c,d)$ together with $(a,b)=(c,d)$ would substitute to $(a,b)\prec (a,b)$, excluded by step 1.1; the same applies to $(c,d)\prec (a,b)$ together with equality; and $(a,b)\prec (c,d)$ together with $(c,d)\prec (a,b)$ is impossible, since the definition of $\prec$ leaves four combinations: $a<c$ with $c<a$; $a<c$ with $c=a$ and $d<b$; $a=c$ and $b<d$ with $c<a$; and $a=c$ and $b<d$ with $c=a$ and $d<b$; the first three contradict trichotomy of the order of $\mathbb{Z}$, and the fourth gives both $b<d$ and $d<b$.

$-1<0$ in $\mathbb{Z}$: adding the integer $-1$ to both sides of $0\le 1$ from step 1.5 gives $0+(-1)\le 1+(-1)$, that is $-1\le 0$; and $-1\mathrm{\ne }0$, because $-1=0$ would give $1=1+0=1+(-1)=0$, contradicting $1\mathrm{\ne }0$; hence $-1<0$.

For every integer $n$ one has $n+(-1)<n$: adding $n$ to $-1\le 0$ gives $n+(-1)\le n+0=n$; and $n+(-1)\mathrm{\ne }n$, because equality would give $-1=(n+(-1))+(-n)=n+(-n)=0$; so $n+(-1)<n$.

No member of $\{0\}\times \mathbb{Z}$ is $\prec$-least: given $(0,n)$ in the set, the integer $n+(-1)$ exists in $\mathbb{Z}$ and $(0,n+(-1))$ again lies in $\{0\}\times \mathbb{Z}$ by step 1.4, with $0=0$ and $n+(-1)<n$ by step 3.1, so $(0,n+(-1))\prec (0,n)$: every member of the subset has a strictly smaller member of the subset below it.

The relation $\prec$ is irreflexive and transitive by steps 1.1 and 1.3, and trichotomous by steps 1.2 and 2.1, so it is the strict presentation of a total order on $\mathbb{Z}\times \mathbb{Z}$; the set $\{0\}\times \mathbb{Z}$ is nonempty by step 1.4 and has no element $\mathrm{\ell }$ with no $s\prec \mathrm{\ell }$, by step 4.1; hence the least-element clause of a well-order fails and $\prec$ is not a well-order on $\mathbb{Z}\times \mathbb{Z}$. The refuted claim is therefore false.