Finite fragments of ZFC

We want to define $E$ over $\mathrm{ZFC}$ such that $\mathrm{ZFC}\vdash\varphi^{\omega,E}$ for every $\varphi\in\mathrm{ZFC}$. If $\mathrm{ZFC}$ is inconsistent in the real world, then it proves everything and we are done. So we may assume $\mathrm{ZFC}$ is consistent in the real world. I think this is the key point of the argument.

Let $\varphi\in\mathrm{ZFC}$. To show that $\mathrm{ZFC}\vdash\varphi^{\omega,E}$, we take an arbitrary model $V\models\mathrm{ZFC}$, and attempt to prove $V\models\varphi^{\omega,E}$. If $V\models\mathrm{Con}(\mathrm{ZFC})$, then we are in the ‘straightforward’ case as you observed. So we may assume $V\models\neg\mathrm{Con}(\mathrm{ZFC})$.

Now $V\models\mathrm{ZFC}+\neg\mathrm{Con}(\mathrm{ZFC})$. Following Kunen's hint, let $n$ be the largest $n\in\omega^V$ such that $V\models\mathrm{Con}(\mathrm{ZFC}_n)$. This $n$ must be bigger than all natural numbers $i$ in the real world because of what Kunen says at the beginning. As a result, if we apply the completeness theorem in $V$ to find a (countable) model $M$ of $\mathrm{ZFC}_n$ in $V$, then $M\models\mathrm{ZFC}_i$ for every natural number $i$ in the real world, which is the same as saying $M\models\mathrm{ZFC}$ in the real world. Notice $V\not\models(M\models\mathrm{ZFC})$ in this case.

Sorry for referring to the ‘real world’ so many times. I cannot find a better phrase for it. Suggestions are welcome.