(This question arose from a homework question which asked me to prove that (1st order) induction is independent from the other (1st order) Peano axioms)
Let $\mathcal{L}$ be the language of Peano arithmetic, i.e. with $0$ the only constant symbol, $+$, $\cdot$, and $S$ (the successor function) the only function symbols, and $=$ the only predicate symbol. Let $A$ be the theory in $\mathcal{L}$ whose axioms are the Peano axioms. Denote by $\overline{n}$ the term $S(S(\cdots S(0)\cdots))$, where $S$ is applied to $n$ times to $0$.
Next, extend $\mathcal{L}$ to a language $\mathcal{L}'$ with a single new constant symbol $\omega$. Let $A_1$ be $A$ extended to $\mathcal{L}'$, and set $$ A'=A_1\cup\{\overline{0}<\omega,\,\overline{1}<\omega,\,\overline{2}<\omega,\,\ldots\}.$$
$A'$ is consistent:
Let $T=A_1\cup\{\overline{i_1}<\omega,\,\overline{i_1}<\omega,\,\ldots,\overline{i_n}<\omega\}$, where $i_1,\ldots,i_n$ are natural numbers (if $n=0$, set $T=A_1$). Let $\mathbb{N}$ be the standard model of $A$. Then $\mathbb{N}$ may be interpreted as a model of $T$ by interpreting $\omega$ to be $\max\{i_0,\ldots,i_n\}+1$. Hence, $T$ is consistent. Thus, by the compactness theorem, $A'$ is consistent.
Question:
Let $M$ be a normal model of $A'$. Then $M$ is also an interpretation of $\mathcal{L}$, and all the axioms of $A$ are true in $M$ since they're also axioms of $A'$. So, $M$ is a model of $A$.
On the other hand, $\vdash_A x\neq0\Rightarrow (\exists y)(x=S(y))$. So, since $M$ is a model of $A$, it follows that $\models_M x\neq0\Rightarrow (\exists y)(x=S(y))$. So, $\omega\in M$ is the successor of some object in $M$. Repeating this, we get a sequence $$\omega>\omega-1>\omega-2>\cdots$$ in $M$ with no minimal element. But this contradicts the well ordering theorem, which is a theorem of $A$!
Where did I mess up?
