Is least upper bound property of the set of real numbers necessary to prove the nested interval theorem ? I know that it is sufficient, but I doubt whether it is necessary.
It is not necessary, in the following sense. There exists an ordered field $S$ where the nested interval theorem holds, but the least upper bound property does not. To prove the nested interval theorem for $S$, one need not assume LUB; indeed one cannot assume LUB, since it isn't true!
As usual for these questions, my source is James Propp's "Real Analysis in Reverse".
It is necessary because the nested interval theorem is not true in $\mathbb Q$. Consider for instance the bisection method for solving $x^2=2$ starting from the interval $[0,2]$. You get a sequence of nested closed intervals with rational endpoints that has empty intersection.