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 $A$.