Prove that the limit of surface areas of the polyhedrons is 4 pi.

Let $P_n$ be a sequence of polyhedrons inscribed into unit sphere or radius $1$, such that maximum size $s$ of all its faces tends to zero $s \rightarrow 0$. Size of the face is maximum distance between any two points of the face.

Prove that limit of surface areas $S$ of polyhedrons is $4 \pi$, that is

$\lim_{s \rightarrow 0} S(P) = 4 \pi$