So, I've lately been having confusion on how to understand infinity, but I think I have progress in my intuition. So, I'd appreciate if someone would let me know if I'm on the right track, and which parts I still don't have right.
This is my understanding so far. Originally I had looked at the infinite intersection: $$\bigcap_{n\in\mathbb{N}} \left(0,\frac{1}{n}\right)$$ I had thought that $\bigcap_{n\in\mathbb{N}} \left(0,\frac{1}{n}\right) \neq \emptyset$ since for an arbitrary $n \in \mathbb{N}$, the intersection is non-empty. Thereofre, since every element in the intersection is non-empty, it must be that the total intersection is non-empty since the intervals are nested.
My current understanding is this: that while the above information could be used to prove that any finite/arbitrarily long intersection is non-empty, an arbitrarily large number will never cover all of $\mathbb{N}$. I reasoned that for any $m \in \mathbb{N}$, $\bigcup_{n=1}^{m}n$ does not cover all the natural numbers, since $m+1 \notin \bigcup_{n=1}^{m}n$. So while it covers arbitrarily many natural numbers, it can't possibly cover all of them.
So the only way to cover all of the natural numbers is using some object which is not in $\mathbb{N}$ and is greater than all natural numbers. If I have an $M$ such that $\forall n \in \mathbb{N},M > n$, then $\mathbb{N} \subset \bigcup_{n=1}^{M}n$. So, even if the intersection is non-empty for each natural number, its still neccesary to check that $(0,\frac{1}{M})$ is non-empty. Checking this, gives that its non-empty only if $\exists n\in \mathbb{N} \text{ such that } 0<\frac{1}{n}<\frac{1}{M}$. Since this implies $M<n$, but $\forall n\in\mathbb{N}, M>n $, then this fails for all natural numbers.
While my explanation is very unrigorous, it seems like it can be formalized by using limit ordinals, and including limit ordinals as a requirement when doing induction. Is this correct? Is this the right way to view these problems? Also, does a value the same size as the limit oridinal occur infinite times, or does it only "occur" at the end? More formally, would $$\left|\bigcup_{n \in\mathbb{N}} \begin{cases} \emptyset, \text{if} \left(0,\frac{1}{n}\right) \neq \emptyset \\\{n\},\text{otherwise} \end{cases} \right|$$ be 0, or $\omega$?
There's a lot to unpack here.
First, there seems to be an issue with a fact of typical induction that (in my experience) sometimes escapes students: a statement being true for each $n\in\Bbb N$ does not mean it 'spills over' to infinity.
Typical induction goes as follows: $P(n)$ is a proposition about $n\in\Bbb N$. If $P(0)$ is true and if $P(n)$ being true implies $P(n+1)$ is true, then $P(n)$ is true for all $n\in\Bbb N$.
Notice how this is fundamentally a result about natural numbers, all of which are finite. Infinity is not a natural number. You yourself have pointed that out, and indeed in mentioning ordinals it seems you might be acquainted with (or would be interested in) transfinite induction. Transfinite induction is what's used when we want a proposition to 'spill over' to infinity, and it tells us precisely that one needs to take special care with this transition (more precisely, the case of limit ordinals needs its own inductive step).
Moreover, and somewhat related, is the fact that, outside the context of induction, but in the context of limits, there are properties that may hold in each term of a limit but fail to hold in the limit.
You might have heard of fluke proofs that $\pi = 4$, by starting from a square that is progressively made to approach the circle that inscribes it. In each step the resulting polygon preserves the perimeter of the initial square, and yet in the limit, we get the circle.
This is an example where the kind of convergence matters. In this case, the limiting process is a pointwise convergence, for which plenty of properties fail to hold over the limit. There are other kinds of stronger convergence (like uniform convergence), for which more properties hold over the limit.
Now, perhaps this example was discrete, and therefore it still smells of typical induction. Of course, rather than consider it to be done in steps, you could smooth out the transition between each step along the real numbers, so the whole thing becomes a continuous limit.
But you can observe this in much simpler cases too. Consider $\lim_{x\to\infty}1/x$. This is a continuous limit, and for every $x>0$ the expression $1/x$ is positive. And yet, the limit is $0$ (non positive).
Finally, there seems to be just plain misunderstanding about what the symbol $\bigcap_{n\in\mathbb{N}}$ means. Or rather, given what you state in your question, it seems your intuition fails to conform to the understanding of the symbol.
There are plenty of counterintuitive things in mathematics. Most paradoxes are usually named so not because they're wrong, but because when they were found out, they were deemed counterintuitive.
It is important to be able to recognize when our intuition fails. This is usually a moment of growth: if we understand a posteriori why something fails, we'll have learned something valuable from it. This applies to both points I shared previously: we learned something about induction, and we learned something about limits.
In the case of $\bigcap_{n\in\mathbb{N}} A_n$, the meaning is the set of elements that belong to all $A_n$. In your example question, $A_n = (0,1/n)$. We have
$$\begin{align} a \in \bigcap_{n\in\mathbb{N}} \left(0,\frac1n\right) &\iff \forall n\in\Bbb N, a \in\left(0,\frac1n\right) \\&\iff \forall n\in\Bbb N, 0 < a < \frac1n \end{align}$$
so $a$ is positive, and yet, $n<1/a$ for all $n\in\Bbb N$. We are forced to conclude that no such $a$ exists, so the intersection $\bigcap_{n\in\mathbb{N}} \left(0,\frac1n\right)$ must be empty.