$\bigcap_{n=1}^\infty A_n$ 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.
One nitpick I can see is that you inferred the intersection is $[a,b]$ by considering limits of the interval endpoints. However, it's not obvious that $a,b\in \cap_{n=0}^\infty I_n$ from the point of view of sets. It becomes clearer if you state that since $[a_{n+1},b_{n+1}]\subseteq [a_n,b_n]$, if $a\notin \cap_nI_n$, then $a\notin I_{k}$ for some $k>0$, which implies either $a<a_k$ or $a>b_k$, both of which will yield contradictions.
As an alternative, you can try to show that $[a,b]\subseteq \cap_{n=0}^\infty I_n$ and $[a,b]\supseteq \cap_{n=0}^\infty I_n$.
Best Answer
$\forall n\in\Bbb N$ never applies to $\infty$, because $\infty$ is not an element of $\Bbb N$. In the nested interval theorem there is no $I_\infty$. What we know is that $x\in I_n$ for each $n\in\Bbb N$, and therefore by definition $n$ is in the intersection of the sets $I_n$. You could call this intersection $I_\infty$ if you wanted to do so, but that would be an arbitrary choice altogether independent of the induction argument involving the sets $I_n$; you could just as well call it George. (Many years ago a friend of mine did in fact publish a paper about a mathematical object that he named George.)
As for De Morgan’s law, one proves it for arbitrary families of sets simply by showing that each side of the proposed identity is a subset of the other. This is done for arbitrary indexed families of sets here and in this answer (and probably other places at MSE as well). The proof does not depend on the theorem for finite families of sets and does not involve any kind of induction.