Nest theorems never explain why nests have to be closed. Does this float?
Given nests (ai,bi), (ai,bi], [ai,bi] and limai=limbi=a. Assume x belongs to all intervals. Then:
lim(ai<x<bi)=(a<x<a) => no x in (ai,bi)
lim(ai<x bi)=(a<x a)=> no x in (ai,bi]
lim(ai x bi)=(a x a) => x=a in [ai,bi]
(-1/n,1/n) no x
(-1/n,1/n] no x
Plato is right. OP sinks. Thanks. A consideration of what went wrong leads to the following:
Open, Closed, and Half-Open Nests
Given nests of intervals ai,bi = (), (], [), or  and limlbi-ail=0 so that limai=limbi=a. When is there a point common to all intervals?
limai=a but no ai=a
limbi=b but no bi=a
Then ai<a<bi, ai<a<=bi, ai<=a<bi, ai<=a<=bi and (), (], [),  all have a common point.
limai=a and ai=a, i>N
limbi=a but no bi=a
Is there an x st ai<x<bi or a<x<=bi i>N? NO. So nests (ai,bi) and (ai,bi] have no common point. But ai<=a<=bi and ai<=b<bi so [ai,bi], [ai,bi) have a common point.
limai=a but no ai=a
limbi=a and bi=a, i>M
Then (ai,bi) and [ai,bi) have no common point, but [ai,bi] and (ai,bi] have a common point. Same reasoning as 2).
There are cases where that concept is very general, see Kelly.
But you seem to restrict consideration to real numbers.
Saying that is a nested sequence of sets usually means that if then .
If that is what nest means then the sequence is increasing, bounded above; the sequence is decreasing , bounded below. In both cases the sequences converge and the limit(s) belong to the common part of the nest.
QUESTION: Do you mean something different by nest?
The closest any text book comes to explaining why the intervals have to be closed in order to have a common point is to give a counter example. However, as I have shown, depending on situation, open and half open intervals can also have a point in common, especially in the fairly general case where ai<a and a<bi all i when they will always have a point in common.
I didn't take the time to describe exactly with math typography what is available in any text under "nested intervals" as it would take me forever to copy it from a text. As I said, I assumed famliarity with nested intervals, and besides, I'm lousy at TEX. I'm sure you can google the classical proof of intersection of closed nested intervals.
edit: The crucial question is nests of real numbers. Nests in n-space then determine real nests of the individual coordinates.
It may help to see exactly what you are asking.
Frankly, I am confused as to your actual question.
If you consider to fit your description, ie
then both sequences converge to one but the intersection is empty. The limit point is not in any of the sets. So again the exact description of a nest makes a difference.
As for my actual question: why do std proofs that the intersection of closed intervals have to be closed never explain why other than to give a counter example?
OK, a counter example occurs to me. The nest (0,1/n) has no common point but [0,1/n] does. This case is covered in the complete theorem on nested intervals in post #6
On the other hand for the limit point belongs to ALL of the sets. So the intersection is not empty. But the sets are not closed either.
So to insure an nonempty intersection the sets in the nest must be closed. You can prove that.
Sorry, but I have been shut out of MHF on my home computers, and its getting crowded here at the public computers, and anyhow, I want to go home. I really thought what I did in post #6 was straight-forward and self-explanatory, assuming familiarity with the std nested interval theorem. At this point, readers will have to judge for themselves, or I will try to answer more questions the next time I come here.
I do appreciate your finding the fault with my OP.
The standard theorem says that, using closed intervals, we can ensure a common point, belongs to all of the nested intervals.
And the counter example of (0,1/n] says, using non-closed nested intervals, we can't ensure there is a common point, belongs to all of them.
That's it. This explains why "closed" is required by the theorem.
I didn't see any problem here.