A bounded closed set may be something else than just a closed interval or a finite union of closed intervals. It can be pretty complicated as well (cf. Cantor sets).

What is the L-2 norm of a set? You can use the diameter of a set: , where is any norm.then we have that the intersection of all I_k's is nonempty. furthermore, if L(I_k) denotes the length of each I_k, and if lim(k->infinite) L(I_k) = 0, then intersection of all I_k's is a singleton.

to prove this theorem in the space R^n would i just consider the L-2 norms of each set?

If you know the 1-dimensional case (not only for intervals), you can prove the general case by considering the projections of the sets on the axes (they are compact as well, i.e. closed and bounded, but this is not a straightforward consequence of being closed and bounded, be careful), and applying the 1-dimensional theorem to these.

Otherwise, it would really help to know that "A is bounded and closed" (in ) is equivalent to (or just implies that) "every sequence in A possesses a converging subsequence". (I think this is called Bolzano-Weierstrass property)

If you know that, choose for every (I assume the sequence of sets is ). Then there is a converging subsequence (because ) and its limit is in , but also in (because ), and so on, so that the limit belongs to the intersection of all the : this intersection is non-empty. If the diameters tend to 0, you easily see that two points in the intersection must be arbitrarily close to each other (at a distance less than the diameter of , which tends to 0), and hence equal.