Let x and y be nonunits in a valuation ring R. If xz is a unit then x would be too, so x times anything in R remains a nonunit. If x+y is also a nonunit we have an ideal.
Since x and y are not units, 1/x and 1/y are not in R. Without loss of generality assume x/y is in R. Hence 1+x/y = (x+y)/y is in R. If 1/(x+y) is also in R then so is 1/y, which is a contradiction. Therefore x+y is not invertible, and the nonunits form an ideal in R.
Call this ideal M, since it is maximal. After all, the only things not in M are units. Note that M contains every other proper ideal. Thus M is the maximal ideal of R, and R is a local ring.
Let x, an element of F-R, be integral over R. Thus x satisfies a monic polynomial p(x). Write xn in terms of smaller powers of x, then divide through by xn-1. Now x is a linear combination of powers of 1/x, with coefficients in R. Remember that 1/x is in R, thus x is in R, which is a contradiction. Thus R is integrally closed in its fraction field.