Let x be integral over the ring R/S. Thus xn is equal to a linear combination of lower powers of x.
Look at the coefficients on these lower powers of x, and let d be the product of their denominators. Then consider (dx)n. This multiplies our expression by dn. The result is a linear combination of lower powers of dx, with coefficients in R. (We need an integral domain here, to make sure the equation doesn't drop to 0 = 0.) Therefore dx is integral over R.
Yet R is integrally closed, so dx lies in R. Since d is the product of elements of S, dx/d lies in R/S. Thus x is in R/S after all, and R/S is integrally closed.
Recall that Z, the ring of integers, is integrally closed. Applying the above, Z localized about any prime p is also integrally closed.