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

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

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

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