Let A={nεN:n>y}.Now that set is not empty because the Archimedean principal tell us that:

for all y>0 there exists nεN SUCH THAT n>y

SINCE A is not empty and also a subset of the natural Nos N, then according to the well ordering principal of natural Nos :

there exists a kεA AND .

But since kεA ,THEN y<k. ALSO k-1<k.

If now y<k-1 ,then (k-1)εN which implies a contradiction ,hence :

For part (b).

Assume that the set A HAS two minimums k,m both belonging to A

But since m is a minimum , ...........(1)

Also since k a minimum , .............(2)

From (1) and(2) we conclude k=m ,therefor k is unique

Do not forget that a minimum of a set is an infemum that belongs to the set.

If you do not understand part (b) i can give a more detailed proof