Your attempted proof only handles uniqueness of $a$, not existence. The uniqueness part can be done much more easily, noting that $\bigcap_{k=1}^\infty F_k\subseteq F_n$ for any $n$, therefore $\diam\Bigl(\bigcap_{k=1}^\infty F_k\Bigr)\le\diam(F_n)$. Since $\diam(F_n)\to0$, this implies $\diam\Bigl(\bigcap_{k=1}^\infty F_k\Bigr)=0$ (provided the intersection is nonempty), so there is only one point in the set. — Harald Hanche-Olsen Oct 9 '13 at 18:22