How can you prove the Nested Interval(s) Theorem in $\mathbb R$ without reference to the least upper bound property? Claim: a nested sequence of closed bounded intervals in $\mathbb R$ has nonempty intersection.
A textbook provides a proof using the least upper bound property of the real numbers, but adds an aside that the theorem can be proved in $\mathbb R$ in alternate ways. I'm hoping for some examples of proofs that do not use the least upper bound property.
 A: As Yuval Filmus says in the comments, an important basic observation is that a nested collection of closed bounded intervals in $\mathbb{Q}$ may have empty intersection. Thus any proof of this property must make use of a property of $\mathbb{R}$ which is not a property of $\mathbb{Q}$. The least upper bound property is one of the most important such properties, and there are others, many of which turn out to be equivalent (in an appropriate sense) to the least upper bound property. 
A: Here is a sense in which the LUB property and NIT property
are mathematically equivalent for the purpose arising in
your context. This theorem provides a sense (namely,
separable linear orders) in which it isn't possible to
prove the NIT in a context where LUB fails; they arrive
together. Thus, the question is one of presentation rather
than mathematical possibility.
Theorem. Suppose that $\langle R,\lt\rangle$ is any
linear order with a countable dense set. Then it satisfies
the LUB property if and only if it satisfies the NIT
property.
Proof. Suppose it satisfies the LUB property, and
$[a_0,b_0]\supset [a_1,b_1]\supset\cdots$ is a descending
sequence of intervals. It follows easily that the least
upper bound of the $a_n$ is in every interval, and so the
NIT property holds. Conversely, suppose that the NIT
property holds and that $A\subset R$ is nonempty and
bounded above. Enumerate the countable dense set
$Q=\{q_n\mid n\in\mathbb{N}\}$, and define a sequence of
intervals $[a_n,b_n]$ as follows. The intervals will zero in on where we expect to find the lub of $A$. Choose any $a_0\in A$ and
any upper bound $b_0$ of $A$ in $R$. If $[a_n,b_n]$ has
been defined, then consider $q_n$. We keep the same
interval, unless $a_n\lt q_n\lt b_n$. In this case, we shrink the interval so as to decide $q_n$. That is, if $A$
has elements from the top half $[q_n,b_n]$, then we shrink
the interval on that side with
$[a_{n+1},b_{n+1}]=[q_n,b_n]$, and otherwise we shrink the
interval towards the other side with
$[a_{n+1},b_{n+1}]=[a_n,q_n]$. One can now easily prove
that $A$ has elements from every interval, that $A$ is
bounded above by every $b_n$, that there is a unique
element in the intersection of the intervals, and that this
element is the least upper bound of $A$. QED
Things become interesting when one drops the
countable-dense-set hypothesis. As the argument shows, the
LUB property still implies the NIT property in any linear
order, but the converse can fail. One easy (counter)example
is the order $\mathbb{R}^\ell\oplus\mathbb{R}$, that is,
the long
line
$\mathbb{R}^\ell$ with a copy of the reals on top; the order
$\omega_1+\omega^\ast$ works just as well. These orders
have the NIT, because any countable sequence of intervals
that eventually stays on one side of the principal cut will
be fine, and those that straddle the cut will have the left
hand sides bounded in $\mathbb{R}^\ell$, since every
countable subset of the long line is bounded, and hence
have nonempty intersection. But the order does not have the
LUB property because of the principal cut between the two
orders.
One might be tempted to extend the NIT to longer
transfinite nested sequences of intervals, but actually
these two orders continue to have the NIT property even for
such longer transfinite sequences. This is because of the
mis-match in cofinality between the lower and upper sides
of the principal cut. If the length of a sequence of
intervals straddling this cut has countable cofinality,
then it will be bounded in the lower part, and if it has
uncountable cofinality, then it will be eventually constant
in the upper part; so in any case, sequences of nested
closed intervals of any transfinite length will have a
nonempty intersection, but the LUB property fails.
The resolution is not to use sequences, but to use filters
(or nets), and this is a standard idea in topology when one
gets away from the separable case.
A linear order has the LUB property if and only if every filter of closed intervals has nonempty intersection. The forward direction is similar to the above, and for the reverse, consider the collection of intervals that straddle where you think the LUB of a set $A$ should be, and the unique point in the intersection of them will be the LUB of $A$.
