Let F be the fraction field of R. Let x, a member of F, lie in the intersection of the localizations of R.
Let J be the ideal of R that maps x into R. In other words, y ∈ J iff yx ∈ R.
If you write x as a fraction, then the denominator lies in J.
If J is all of R then J contains 1, and x is already in R, so assume J is a proper ideal of R. Thus J is contained in a maximal ideal M in R.
If x can be represented by the fraction a/b, then b lies in J, which lies in M.
Localize about M, and suppose x is represented by some a/b in RM. We just showed b lies in M, hence a/b is not part of RM. This is a contradiction, hence x lies in R after all, and the intersection of localizations resurrects R.