Have you proved yet that there does exist an integer n, such that ? For this kind of problem, it is generally hard to know exactly what you may and may not use to solve it.
By the Archimedean property, there exists an integer n, such that x<n. Hence the set is finite. Let n be the largest integer in this set. Then , but x<n+1, because otherwise n+1 would be in the set, contradicting maximality of n. This show existence.
Suppose both m and n have the desired property, and suppose contrariwise that m<n. Then , and so
contradicting the choice of m