I enumerate the axioms of $\mathbb{ZFC}$ as $\phi_{n} : n \in \mathbb{N}$ in such a way that each axiom appears infinitely often in the enumeration. Then I recursively construct a strictly increasing sequence of ordinals {$\beta_{n} : n \in \mathbb{N}$} such that each subformula of $\phi_n$ is absolute for $V_{\beta_{n}}$.
Let $\delta_0$ = sup{$n : n \in \mathbb{N}$}.
I know that $V_{\beta_{0}} \models \phi$
For each axiom of ZFC, so we know that there is ordinal $\delta$ s.t. $V_{\delta} \models$ ZFC
Now, since each axiom of ZFC holds in $V_{\delta_{0}}$ so too does their logical consequence, that is:
$$(\exists\delta\in\mathbb{ON}. V_{\delta}\models ZFC)^{V_{\delta_{0}}}$$
So there is some $\delta_1 \in \delta_2$ such that $V_(\delta_1) \models ZFC$
Again, since each axiom of ZFC holds in $V_(\delta_1)$ so too does their logical consequence:
$$(\exists\delta\in\mathbb{ON}. V_{\delta}\models ZFC)^{V_{\delta_{1}}}$$
We obtain $\delta_2 \in \delta_1$ such that $V_(\delta_2) \models ZFC$. We can continue in the same manner, obtaining { n : n 2 N}. This is an infinite decreasing sequence of ordinals.
What is wrong with this proof?
Here's the key error. You can't construct such a sequence (at least not using only ZFC). For each individual formula $\phi$, ZFC proves that you can find $V_\beta$ which is absolute for $\phi$. But this is a theorem scheme: for each formula $\phi$, there is a separate theorem of ZFC saying you can do this. There is no single theorem in which $\phi$ appears as a universally quantified variable (represented by a Gödel number, say) which says "for all $\phi$, there exists $V_\beta$ which is absolute for $\phi$". Indeed, you can't even express what "$V_\beta$ is absolute for $\phi$" means when $\phi$ is represented as a Gödel number, since truth in $V$ is not definable.
Without a single general such theorem, you cannot make a recursive definition of the sequence $(\beta_n)$, and so the argument fails at the start.