Can we take the intersection of ALL successor(inductive) sets?

In Halmos Naive set theory, there is the following excerpt introducing natural numbers :

In this language the axiom of infinity simply says that there exists a successor [inductive] set A. Since the intersection of every (non-empty) family of successor sets is a successor set itself (proof?), the intersection of all the successor sets included in A is a successor set $$\omega$$.

I can prove the first part(given that the family is finite), however, I'm not sure why we can extend the conclusion to the intersection of all the successor sets(bolded), which may be an intersection of infinitely many sets, due to the Axiom of Infinity. However, in Halmos book infinite intersection is not yet defined... To be succinct, my problem is focused on the word "all", why it could be used here?

• See page 15 for the definition of the intersection $\cap \mathcal C$ of a (non empty) collection $\mathcal C$ of sets : there is no restrictions, provided that the collection is not empty. – Mauro ALLEGRANZA Jan 28 at 8:08
• Thus, the argument of page 44 is : we have a successor set by Axiom of Infinity. Thus, consider the collection of all successor sstes (that is not empty by the axiom) and apply the operation of intersection above. – Mauro ALLEGRANZA Jan 28 at 8:11
• @MauroALLEGRANZA Thank you. I checked that page and persuaded myself indeed there's no limitation on the "finiteness" of the collection for intersection. I should have read the text more carefully. Please, add this comment as an answer and I should accept it, as it's the crux of my question here. – Macrophage Jan 28 at 8:17

The key point is the point right after your bold text. Included in $$A$$. Simply consider
$$\{a\in A\mid\forall B\subseteq A:B\text{ is a successor set}\to a\in B\},$$
Or, in simpler terms, $$\bigcap\{B\subseteq A\mid B\text{ is a successor set}\}$$. Separation is enough here, of course.
What is important for this proof, as a whole, is that $$A$$ is any successor set. It does not matter which one you take, so as long as you have at least one of them, you define the same $$\omega$$.