I am not quite sure how to prove this. Well, if $\mathfrak{N}$ is the standard model of arithmetic then by definition, a nonstandard model of arithmetic $\mathfrak{N}'$ must be elementarily equivalent to $\mathfrak{N}$ but not isomorphic to $\mathfrak{N}$.
The sentence $(\forall x)( \exists y)(x<y)$ is true in $\mathfrak{N}$ and so it must be true in $\mathfrak{N}'$ as well. But then how will I find a element $b$ in the universe of $\mathfrak{N}'$ such that $a<^{\mathfrak{N}'}b$ for each $a$ in the universe of $\mathfrak{N}'$?
I'm not sure how to proceed. Hints will be appreciated.
Here's my attempt after receiving hints from Asaf Karagila and Rob Arthan
:
Let $N$ be the set of Peano Axioms (see Example 2.8.3). Let $\mathfrak{N}$ be the standard model of arithmetic and $\mathfrak{N}'$ be a nonstandard model of arithmetic. Let $\overline{n}$ be the $\mathcal{L}_{NT}$ term given by $\overline{n}=\underbrace{SS\ldots S}_{n \text{ many}}0$. Let $\Sigma=\{ x=0 \vee x=\overline{1} \vee\ldots\vee x=\overline{n} \vee \overline{n}<x : n\in \mathbb{Z}^{+}\}$. Let $A$ be the universe of $\mathfrak{N}'$
Claim 1. . For each $n\in\mathbb{Z}^{+}$, $N\vdash x<\overline{n} \leftrightarrow x=0 \vee x=\overline{1} \vee \ldots \vee x=\overline{n-1} $.
Proof of Claim 1. We proceed by induction on $n$. Consider the case when $n=1$. We have to prove that $N\vdash x<S0\leftrightarrow x=0$. To prove this, we use Quantifier Axioms and 10th Axiom of Peano Axioms (see Example 2.8.3). We have $N\vdash x<S0 \leftrightarrow x<0 \vee x=0$. But then by the 9th Axiom, $N\vdash \neg x<0$. So by Propositional Consequence, $N\vdash x<S0 \leftrightarrow x=0$ which is what we wanted.
Assume that we already have $N\vdash x<\overline{n} \leftrightarrow x=0 \vee x=\overline{1} \vee \ldots \vee \overline{n-1} $ for some $n\in \mathbb{Z}^{+}$. Let's prove that $N\vdash x<\overline{n+1} \leftrightarrow x=0 \vee x=\overline{1} \vee \ldots \vee x=\overline{n} $.
Now, $N\vdash x<\overline{n+1}\leftrightarrow x=\overline{n} \vee x<\overline{n}$ by the 10th Axiom and Quantifier Axiom. By Propositional Consequence and Induction hypothesis, we have what we wanted to prove.
Claim 2. For each $\sigma \in \Sigma$, $N \vdash \sigma$.
Proof of Claim 2. Let $n\in \mathbb{Z}^{+}$. We show that $N\vdash x=0 \vee x=\overline{1} \vee\ldots\vee x=\overline{n} \vee \overline{n}<x$. By Axiom 11 and Quantifier Axioms, $N \vdash x<\overline{n} \vee x =\overline{n} \vee \overline{n}<x$. It is easy to see that Claim 1 and Propositional Consequence complete the proof of Claim 2.
Claim 3. $A\setminus \{ \underbrace{S^{\mathfrak{N}'}(S^{\mathfrak{N}'}(\ldots S^{\mathfrak{N}'}}_{n \text{ times}}(0^{\mathfrak{N}'})\ldots)) : n \ge 0 \}$ is nonempty for otherwise $\mathfrak{N}\simeq \mathfrak{N}'$.
(Proof of Claim 3 is pretty long but straightforward, so, I will not prove it here).
Now, $\mathfrak{N}\models \Sigma$ and since $\mathfrak{N}'$ is elementarily equivalent to $\mathfrak{N}$, we have $\mathfrak{N}'\models \Sigma$. Let $a$ be any nonstandard element of $\mathfrak{N}'$ (Claim 3 proves the existence of such an element). Let $s$ be any variable assignment function into $\mathfrak{N}'$ which sends $x$ to $a$.
Then for each $n\in \mathbb{N}$, $\mathfrak{N}' \models x=0 \vee x=\overline{1} \vee\ldots\vee x=\overline{n} \vee \overline{n}<x$. So it must be that $\underbrace{S^{\mathfrak{N}'}(S^{\mathfrak{N}'}(\ldots S^{\mathfrak{N}'}}_{n \text{ times}}(0^{\mathfrak{N}'})\ldots)) <^{\mathfrak{N}'} a$ for each $n\in \mathbb{N}$.
Best Answer
First of all, normally when we say "model of arithmetic" we mean a model of a theory which describes arithmetic, e.g. Peano, rather than true arithmetic.
Secondly, note that an "infinite element" is not $0$, or $s0$, or $ss0$, or $sss0$, or $ssss0$, or $sssss0$, and so on and so on (here $s$ is the successor operation).
Suppose that a model does not contain an infinite element. Well, it must contain $0$ and $s0$ and $ss0$ and $sss0$ and $ssss0$ and so on and so on. But we just said, that it doesn't contain any other elements. So using this, you can now show that it is isomorphic to the standard model.
(Note that your approach is inherently flawed: the statement "$x$ is a non-standard integer" is not first-order definable! It is a property of the model, not an internal property of the element.)