AST is conservative over ZF So this statement that AST is conservative over ZF is an exercise in Ralf Schindler's book, which I need to prove, but I am having trouble with it.
So the formulation of AST(Ackerman's Set Theory) which I will be using, is the following: Our language is $\mathcal{L}_\epsilon$ together with a constant symbol $\dot{v}$. And the axioms are Extensionality, Foundation, Comprehension(the same old ZF one), an axiom saying $\dot{v}$ is both transitive and supertransitive and one reflection axiom for each formula $\varphi(v_1, \dots, v_n)$ of $\mathcal{L}_\epsilon$ stating that:
$$\forall v_1\in\dot{v}\dots \forall v_n\in\dot{v} (\varphi^{\dot{v}} \leftrightarrow \varphi)$$
So in a previous exercise I had proven that AST $\vdash$ ZF. So by reflection, we have that $\dot{v} \models $ ZF. Also since AST $\vdash$ ZF, by using the coding in Devlin's book for example, we can see that the satisfaction relation for sets can be constructed in AST, thus AST proves Con(ZF), since $\dot{v}$ is just another set in AST's universe. 
This is my problem.  How can AST be conservative over ZF and also in the mean time prove Con(ZF)?
 A: 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.

A: (I had deleted this answer as it does not answer the question. I undeleted it for the context of the comments in Noah Schweber's answer, with some modifications.)
I checked Reinhardt's original article, and I found that what he proved is the following statement:

Theorem (Reinhardt, Results 0.2.) If $\phi$ is a theorem of $\mathsf{ZF}$, then $\mathsf{AST}$ proves $\phi^{\dot{v}}$. Moreover, the converse also holds.

Hence I think Schindler states the exercise problem somewhat misleadingly. 
I was not correct at that point. OP's comment on the reflection argument seems to work.
I reached an unexpected conclusion: $\mathsf{AST}$ is not $\omega$-consistent$\omega$-complete by the following argument: we may identify formulas to natural numbers by coding. We know that $\mathsf{AST}\vdash \phi\in\ulcorner\mathsf{ZF}\urcorner\to c\models \phi$ for each 'standard' $\phi$. 
If $\mathsf{AST}$ is $\omega$-complete, then we have $\mathsf{AST}\vdash \forall\phi:\phi\in\ulcorner\mathsf{ZF}\urcorner\to c\models \phi$ for each 'standard' $\phi$. This means $\mathsf{AST}\vdash \dot{v}\models\mathsf{ZF}$. 
However, we know that $\mathsf{AST}$ and $\mathsf{ZF}$ are equiconsistent.
Here is another example of an extension of $\mathsf{ZF}$ that is not $\omega$-complete (maybe due to Feferman.) Consider the language $\langle \in,c\rangle$, where $c$ is a constant symbol. Let $T$ be the theory given by the following axioms: the usual axioms of $\mathsf{ZF}$, the statement that $c$ is a transitive set, and $\phi^c$ for all axioms $\phi$ of $\mathsf{ZF}$.
By the similar argument for $\mathsf{AST}$, $T$ is $\omega$-incomplete.
