How to evaluate infinite intersections or unions on any intervals? I'm having trouble understanding the basic principles of evaluating infinite intersections or unions on any intervals. Such as
$$\bigcap_{i=1}^{\infty} [2+i^{-2}, 5-i^{-2}]$$
I've been using the following:
https://math.stackexchange.com/a/1438760/248602
as an example, but I have trouble adapting this to infinite intersections since the arguments should be slightly different.
Can anyone explain the basic rules / common techniques for evaluating infinite intersections or unions for any intervals?

"Prove" is the wrong word.  Maybe you meant "evaluate" or "simplify" or express in some simple form. One proves theorems; one proves identities; one proves propositions, lemmas, corollaries, etc.; one prives things that can be true of false. One does not "prove" intersections of sets.
As it stands now, the expression you wrote is $$\bigcap_{i=1}^\infty \{2+i^{-2}, 5-i^{-2}\}.$$ Maybe you meant $$\bigcap_{i=1}^\infty (2+i^{-2}, 5-i^{-2})$$ or $$\bigcap_{i=1}^\infty [2+i^{-2}, 5-i^{-2}].$$
In conventional notation $\{a,b\}$ means a set with only two members $a$ and $b$, and $(a,b)$ means an open interval from $a$ to $b$, with infinitely many members (except when that notation refers to an ordered pair, which is quite a different thing), and $[a,b]$ means the closed interval from $a$ to $b$.
Let us suppose for the moment that you meant $$\bigcap_{i=1}^\infty (2+i^{-2}, 5-i^{-2}).$$
The main thing you need to know is this: A number (or other object) is a member of the intersection of a specified collection of sets if and only if it is a member of every one of them.
You're looking at
$$
(3,4) \cap (2+\tfrac 1 4, 5-\tfrac 1 4) \cap (2+\tfrac 1 9, 5 - \tfrac 1 9) \cap \cdots.
$$
These intervals are actually getting longer, so the first one is a subset of all of the others.  Therefore the only numbers that are in every one of these intervals are those that are in the first one, $(3,4)$, so that is the intersection.
I am suspicious: I wonder if you saw $\displaystyle \bigcap_{i=1}^\infty (2-i^{-2}, 5 + i^{-2})$ and interchange the plus and minus signs.  In that case, you'd have
$$
(1,6) \cap (2-\tfrac 1 4, 5 + \tfrac 1 4) \cap (2-\tfrac 1 9, 5+\tfrac 1 9) \cap \cdots.
$$
In that case, the lower bound of each interval is less than $2$ and the upper bound is more than $5$.  $2$ and $5$ are both members of every one of those intervals, and so are all numbers between $2$ and $5$. But every number less than $2$ will get excluded by some of the intervals, as will every number more than $5$.  Thus the intersection in that case is $[2,5]$, a closed interval that includes both endpoints.
