To find the subsequence, notice that is the fractional part of , and therefore (by induction) is the fractional part of . It will be sufficient to show that we can find multiples of a with fractional parts arbitrarily close to 0.
Divide the unit interval into k subintervals of length 1/k. The fractional parts of na are all distinct (otherwise a would be rational), and at least one of the subintervals must contain infinitely many of them. So we can find arbitrarily large m and n (with n>m) for which the fractional parts of ma and na differ by less than 1/k. Then the fractional part of (n-m)a is either less than 1/k, or it is greater than 1-(1/k) in which case the fractional parts of r(n-m)a (for r=1,2,3,...) will move down through the unit interval in steps of less than 1/k, until one of them is less than 1/k.