What is wrong with this proof that inaccessible cardinals are inconsistent Disclaimer: I believe this proof is wrong, but I'm asking because I can't find what's wrong with it, which means I must have some basic misunderstanding of the concepts involved.
First, some definitions. Recursively, we define an ordinal $\alpha$ to be $\eta$-unboundedly elementary:
$\alpha$ is $0$-unboundedly elementary iff $\alpha$ is an ordinal
$\alpha$ is $\eta$-unboundedly elementary iff for all $\beta<\eta$, for all $\gamma$, there exists $\delta>\gamma$, such that $V_\alpha\prec V_\delta$ and $\delta$ is $\beta$-unboundedly elementary. That is, $V_\alpha$ is elementary in unboundedly many $V_\delta$ for $\delta$'s lower in this hierarchy.
Some examples: $1$-unboundedly elementary ordinals are those $\alpha$'s such that $V_\alpha\prec V_\beta$ for unboundedly many ordinals $\beta$. In this blogpost they are called totally otherworldly cardinals. $2$-unboundedly elementary ordinals are those $\alpha$'s such that $V_\alpha$ is elementary in unboundedly many totally otherworldly cardinals.
An interesting feature of these unboundedly elementaries is that they exhibit increasing degree of correctness (where $\kappa$ is $\Sigma_n$-correct iff $V_\kappa\prec_{\Sigma_n}V$). Let us first see that $1$-unboundedly elementary ordinals are $\Sigma_3$-correct:
That they are $\Sigma_2$ correct is proven in the blogpost linked above. To see that they are in fact $\Sigma_3$-correct, suppose $\exists x~\varphi(x)$ is a true $\Sigma_3$ statement with parameters in $V_\alpha$. Fix a witness $a$ such that $\varphi(a)$ is true and fix $\beta$ large enough with $a\in V_\beta$ and $V_\alpha\prec V_\beta$. Then since $\beta$ must be a $\beth$-fixed point (this follows from the fact that, for any ordinals $\alpha,\beta$, $V_\alpha\prec V_\beta$ implies $V_\beta\vDash \mathsf{ZFC}$), $V_\beta\prec_{\Sigma_1}V$. Then the truth of the $\Pi_2$ statement $\varphi(a)$ is preserveed downward to $V_\beta$. This means $V_\beta\vDash \exists x~\varphi(x)$, and by elementarity, so does $V_\alpha$. This shows that $\alpha$ is $\Sigma_3$-correct.
Essentially the same trick can give us that $2$-unboundedly elementary ordinals are $\Sigma_4$ correct: this is because there are unboundedly many $\Sigma_3$-correct ordinals that they are elementary in, and if $\exists x~\varphi(x)$ is a true $\Sigma_4$ statement, then we can find some target $V_\beta$ large enough to include a witness, then proceed the same way as in the last paragraph.
Now here's the problem: the recursive definition of $\eta$-unboundedly elementary ordinals is $\Pi_3$. This means ``there exists a $2$-unboundedly elementary ordinal" a $\Sigma_4$ statement. So the least $2$-unboundedly elementary ordinal must be below the least $\Sigma_4$-correct cardinal. But this contradicts the fact we've just shown, that $2$-unboundedly elementary ordinals are $\Sigma_4$-correct!
Nevertheless, assuming there is an inaccessible cardinal, then the usual Löwenheim–Skolem argument shows the consistency of essentially any degree of unboundedly elementary ordinals, and in particular the consistency of $2$-unboundedly elementary.
Taking stock, we have: $\mathsf{ZFC}\vdash ``\text{there exists an inaccessible cardinal}"\rightarrow$ Con($2$-unboundedly elementary), and also $\mathsf{ZFC}\vdash``2\text{-unboundedly elementary ordinals are inconsistent}"$. Putting things together, this shows that $\mathsf{ZFC}$ refutes the existence of inaccessible cardinals. What went wrong?
Some possibilities I can think of: 1) the proof of their increasing degree of correctness is wrong, 2) the complexity calculation of the recursive definition wrong (or maybe it's not recursive after all, that there is some metamathematical issues), 3) inaccessible cardinals doesn't show that these unboundedly elementary ordinals are consistent.
 A: Your claim that "is $n$-unboundedly elementary" is $\Pi_3$ for every finite $n$ (since we already run into issues when $n=2$, there's no point in going transfinite yet) is unjustified: the complexity of the definition as given increases as $n$ does. In general, writing "$E_n$" for "is $n$-unboundedly elementary," we have the suggestive relative complexity expression $$E_{n+1}=\Pi_2(\Pi_1\wedge E_n).$$ The outer $\Pi_2$ is "$\forall b<n\forall \gamma\exists \delta>\gamma\exists W$." This $W$ is implicitly invoked in your post - it's intended to be the pair $\langle V_\alpha,V_\beta\rangle$, and the "$\Pi_1$" in the parenthetical part takes care of saying this. Finally, we have the reference to $E_n$. In general, for $n>0$ the given definition of $E_n$ is $\Pi_{1+2n}$.
The crucial point is that when we invoke $E_n$ in defining $E_{n+1}$, we're doing so within the scope of some already-introduced quantifiers (the "$\Pi_2$" bit). So we don't get this for free.
Note that this is exactly the same reason that the recursive definition of truth doesn't give a contradiction to Tarski's undefinability theorem.
A: The recursive definition of $\eta$-unbounded elementary which you propose cannot actually be carried out in ZFC.  Recall that the way recursive definitions work is that you prove by induction on $\eta$ that there is a unique function $F_\eta$ defined on $\eta$ which takes each $\alpha<\eta$ to the $\alpha$th stage of your definition.  This only works if the $\alpha$th stage of your definition is a set, so that this function $F_\eta$ actually is a set (so you can quantify over such functions).  In your case, you would want to be defining the class of all $\alpha$-unbounded elementary ordinals as the $\alpha$th stage, which you cannot do.  (Note that for variant notions like that of $\eta$-inaccessibility, you can define the set of $\eta$-inaccessible cardinals less than $\kappa$ for any fixed $\kappa$ recursively, since $\eta$-inaccessibility of a cardinal depends only on the cardinals smaller than it.  This doesn't work for $\eta$-unbounded elementary ordinals, though, since to determine whether an ordinal is $\eta$-unbounded elementary, you need to know the entire proper class of $\beta$-unbounded elementary ordinals for each $\beta<\eta$.)
Incidentally, you can carry out this definition in MK, where class quantification is allowed (and can be used in Separation, and thus in proofs by induction where you want to define "the set of counterexamples" to prove it is empty).  However, your $\Sigma_4$-correctness statement then only works for statements using set quantifiers, and the recursive definition of $2$-unboundedly elementary would need to use class quantifiers (or, if you did it with set quantifiers by brute force instead of by recursion, it would need to have higher complexity, as described in Noah's answer).
