Let's give a quick proof. I'll be using the Bolzano-Weierstrass theorem, as well as the fact that continuous functions preserve limits.

First, boundedness. If f weren't bounded above, then for any natural number n we could always find a number xn in the interval with f(xn)>n. By the Bolzano Weierstrass theorem, a subsequence of xn converges to some number x in the closed bounded interval. By construction, f(xn) tends to infinity, and so so does the subsequence in question. But since continuous functions preserve limits, this is a contradiction (we'd have f(x)=infinity). Similarly for boundedness below.

So, f is bounded. Let M=sup f, so f(x) is at most M for all x in the interval. On the other hand, because M is the least upper bound, any number less than M will be exceeded by some value of f. So, for all n, there exists xn with M-1/n<f(xn)<=M. Again by the Bolzano Weierstrass theorem, some subsequence of xn converges to x, and by design, f(xn) converges to M, and then so does the subsequence in question. Since continuous functions preserve limits, we have f(x)=M. Again, lower bounds are similar.