Existence of a first-order set theory corresponding to the first fully correct cardinal

Question

Let $\mathcal{L}$ denote the language of first-order set theory described in Chapter $1$ of “An introduction to set theory” [William A. R. Weiss | October 2, 2008].
Question: does there exist a consistent first-order set theory $T$ that satisfies both of the following two properties?

1. $T$ is infinitely axiomatizable and each axiom is formulated in $\mathcal{L}$ (assuming that the complexity of a formula is not restricted and every formula has a finite length);
2. There exists an ordinal $\alpha$ such that $V_{\alpha} \models T$; the smallest such ordinal is greater than or equal to the initial ordinal of the first fully correct cardinal.

This answer on Mathoverflow may be relevant, but I am unable to find a direct answer to my question.

Definition of the initial ordinal of a cardinal is given in this article.

The smallest fully correct cardinal is the cardinal $\Delta$ described in this answer on Mathoverflow.

Answer 1

If I understand the question correctly, it boils down to the following:

$(*)\quad$ Can there be an $\alpha>\Delta$, where $\Delta$ is$^1$ the supremum of the (parameter-freely, first-order) definable ordinals in $V$, such that $V_\alpha\not\equiv V_\beta$ for any $\beta<\alpha$?

The answer to $(*)$ is, in fact, negative. Say that an ordinal $\alpha$ is fresh iff $V_\alpha\not\equiv V_\beta$ for any $\beta<\alpha$. Freshness is definable, but this means that "the supremum of the fresh ordinals" is a definable ordinal, and hence less than $\Delta$.

Consequently, we have:

For every $\alpha\ge\Delta$ there is some $\beta<\alpha$ with $V_\alpha\equiv V_\beta$.

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$^1$Note that I'm assuming here that $\Delta$ in fact exists. While at first glance a pretty banal statement for "Platonic $V$," there is some surprising nuance here; that said, in the context of this question I think it's worth largely taking for granted.