The formal statement of the principle of mathematical induction is to start with a statement $P(n)$ that depends on a natural number variable $n$. If:
- $P(1)$ is true
- $P(k+1)$ is true whenever $P(k)$ is true.
Then $P(n)$ is true for all $n$.
Let $A_1, A_2, \dots, A_n, \dots$ be an infinite sequence of countable sets. To use induction, you would need to find a statement $P(n)$ such that
$$
P(n) \text{ is true }\forall n \leftrightarrow \bigcup_{n=1}^\infty A_n \text{ is countable}
$$
The statement on the left is that something holds for every number $n$, while the statement on the right is about something accumulated over all $n$.
$\bigcap_{n=1}^\infty A_n$ is a set. What set? The set of all things that belong to every one of the sets $A_n$ for $n\in\Bbb Z^+$. Let $\mathscr{A}=\{A_n:n\in\Bbb Z^+\}$; then $\bigcap\mathscr{A}$ means exactly the same thing. $\bigcap_{n=1}^\infty A_n$ is simply a customary notation that means neither more nor less than $\bigcap_{n\ge 1}A_n$, $\bigcap\mathscr{A}$, and $\bigcap\{A_n:n\in\Bbb Z^+\}$. There is no $A_\infty$: the $\infty$ is just a signal that the index $n$ is to assume all positive integer values.
Suppose that for each positive real number $x$ I let $I_x$ be the open interval $(-x,x)$. Then $\bigcap_{x\in\Bbb R^+}I_x$ is the set of all real numbers that belong to every one of these open intervals. If $\mathscr{I}=\{I_x:x\in\Bbb R^+\}$, then
$$\bigcap\mathscr{I}=\bigcap_{x\in\Bbb R^+}I_x=\bigcap_{x\in\Bbb R^+}(-x,x)=\{0\}\,.$$
How do I know? If $y\in\Bbb R\setminus\{0\}$, then $y\notin(-|y|,|y|)=I_{|y|}$, so there is at least one member of $\mathscr{I}$ that does not contain $y$, and therefore by definition $y$ is not in the intersection of the sets in the family $\mathscr{I}$. On the other hand, $0\in(-x,x)=I_x$ for every $x\in\Bbb R^+$, so $0$ is in the intersection $\bigcap\mathscr{I}$.
In neither case have we used induction anywhere. In the case of the sets $A_n$ we might be able to use induction on $n$ to show that each of the sets $A_n$ has some property $P$, but we could not extend that induction to show that $\bigcap\mathscr{A}$ has $P$. We might somehow be able to use the fact that each $A_n$ has property $P$ to show that $\bigcap\mathscr{A}$ also has $P$, but that would require a separate argument; it would not be part of the induction. The induction argument in that case would prove that
$$\forall n\in\Bbb Z^+(A_n\text{ has property }P)\,;$$
the separate argument would then show, using that result and other facts, that the single set $\bigcap\mathscr{A}$ has property $P$. You could call this set $A_\infty$ if you wished to do so, but that would just be a label; you could equally well call it $A$, or $X$, or even $A_{-1}$, though offhand I can’t imagine why you’d want to use that last label.
In the case of the sets $I_x$ there is no possibility of using induction to show that each $I_x$ has some property: these sets cannot be listed as $I_1,I_2,I_3$, and so on, because there are uncountably many of them. We can still prove things about the set $\bigcap\mathscr{I}$, however. And we could give it any convenient label. $\bigcap\mathscr{I}$ is informative but perhaps a little inconvenient; I might choose to give it the handier label $I$.
In the case of $\mathscr{A}$ there happens to be a customary notation that uses the symbol $\infty$, but that is simply a consequence of the fact that the sets $A_n$ are indexed by integers. We’re doing exactly the same sort of thing in the example with $\mathscr{I}$, but in that case there is no possibility of using a limit of $\infty$ on the intersection, because there is no way to index the uncountably many sets $I_x$ by integers.
Best Answer
The proof that $\bigcap_{n\ge 1}A_n=\varnothing$ when $A_n=\Bbb N\cap[n,\to)$ doesn’t use induction at all. It’s clear that the intersection is a subset of $\Bbb N$, since each of the sets $A_n$ is, so to show that the intersection is empty, we need only show that it contains no element of $\Bbb N$. And that follows immediately from the definition of intersection: for any $k\in\Bbb N$, $k\notin A_{k+1}$, so $k\notin\bigcap_{n\ge 1}A_n$.
By the way, strong induction is equivalent to ordinary induction: anything that can be proved using one can be proved using the other, though converting a proof by strong induction to a proof by ordinary induction typically makes it a little more complicated.