you have two cases for the p, if p is a positive integer you can take the open interval (p-1/2,p+1/2 ) which dose not contains any other element of M except p
if p is not in M, p located between two positive integers say n, n+1 now let "t" be the minimum of |p - n| , |p - (n+1)| so the interval (p-t , p+t) contains p and intersect with M at p just
if p is less than 1 the interval (- infinity , 1 ) contains p and dose not intersect with M.
why you consider just the open intervals with positive integers as a bound ?Solution: Assume that $\exists$ a limit point p of M. Then all open intervals containing p must contain an element of M different from p. Consider the interval I = (s, q) where s \in M and q \in M and s < p < q. That means \exists a point $r \in M$ such that s < r < q. But if q = s+1, then no such r exists because any point between s and s+1 is not an integer. Thus, not every open interval containing p contains a point of M different from p, therefore no limit point of M exists.