Why in Cantor's Intersection Theorem the infinite intersection must contain only one point?

Question

I am going through GF Simmons' Intro to Topology and Modern Analysis.
$F_n$ is a decreasing sequence of non-empty closed subsets of the Metric Space.
It seems that the condition $d(F_n)$->$0$ given in the hyothesis, ensures that the $$F=bigcap_{n=1}^infty F_nneqvarnothing $$ rather F contains exactly one point is not very clear.
Why does "the supremum of the distances between any two points in the closed sets" converging to "$0$"  implies that the set contains exactly one point?

Krishan · Answer

It is because the $$d(A)=sup{d(x,y)|x,yin A }$$
If this converges to $0$ it means that d(x,y) converges to $0$ for the class of closed sets $F_n$ as n->$infty$.
We know that $d(x,y)=0$ only for x=y.Therefore there must exist only one point in the set $F=bigcap_{n=1}^infty F_n$.

Why in Cantor's Intersection Theorem the infinite intersection must contain only one point?

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