AST is conservative over ZF

The accepted answer is incorrect. $\omega$-consistency does not say "If you prove $P(n)$ for each standard $n$, then you prove $\forall n(P(n))$." That would be $\omega$-completeness.

Rather, $\omega$-consistency says "If you prove $\exists x\neg P(x)$, then you must not prove $P(n)$ for each standard $n$." And in particular, both $AST$ and the theory $T$ introduced in that answer are $\omega$-consistent, at least assuming $ZFC$ is to begin with.

Getting back to the main question, it may help to drop $AST$ and consider the simpler fact about $ZFC$ alone, which is itself provable in ZFC:

$(*)\quad$ For every $M\models ZFC$ there is some structure $A\in M$ such that $A\models ZFC$ ... even if $M\models\neg Con(ZFC)$.

The key point is that "$A\models ZFC$" is interpreted in reality; we may not have $M\models(A\models ZFC)$.

This also explains why the OP's reflection argument breaks down - it's exactly the same reason.

Here's how to prove $(*)$ in ZFC:

• If $ZFC$ is inconsistent then $(*)$ is vacuously true.

• Suppose $ZFC$ is consistent. Let $M\models ZFC$. If $M\models Con(ZFC)$ then since $ZFC$ proves the completeness theorem we're done.

• So suppose $M\models \neg Con(ZFC)$. Let $n\in\omega^M$ be what $M$ thinks is the largest number such that there is no proof of a contradiction from the first $n$ axioms of $ZFC$. If we can conclude that $n$ is nonstandard, we'll be done: by completeness in $M$, any model $A$ of the first $n-1$ axioms of $ZFC$ in the sense of $M$ will in reality be a model of $ZFC$, even if $M$ doesn't think so.

Now here's the cute bit: we internalize the reflection principle. Looking at the usual argument we see in fact that ZFC proves "ZFC proves every finite subtheory of ZFC." (Note the crucial nested "proves" here.) This means we can next say...

• Since $M\models ZFC$, for each standard $k$ we have $M\models$ "The first $k$ axioms of $ZFC$ are consistent." So $n$ is nonstandard and we're done.