Several nineteenth-century mathematicians used, without proof, a principle in
their proofs that has come to be known as the nested interval property:
Given a sequence of closed intervals
$$
\left[a_{1}, b_{1}\right] \supset\left[a_{2}, b_{2}\right] \supset\left[a_{3},
b_{3}\right] \supset \ldots
$$
arranged so that each interval is a subinterval of the one preceding it and so
that the lengths of the intervals shrink to zero, then there is exactly one
point that belongs to every interval of the sequence.
Prove this statement. Would it be true for a descending sequence of open
intervals
$$
\left(a_{1}, b_{1}\right) \supset\left(a_{2}, b_{2}\right) \supset\left(a_{3},
b_{3}\right) \supset \ldots ?
$$