Since you haven't showed us what you have done, i will simply give you an outline of the proof.
You probably do realize that in order to show that a composite integer n>1, has a prime divisor p less than sqrt{n}, it suffices to show that n has a divisor, d, less than sqrt{n}.
It is clear that, since n is a composite, then it must have a divisor, call it d', different from 1 and n itself. Hence 2<=d'<n.
Now you will need to consider two cases:
1. If d'<sqrt{n}, then what? and
2. If d'>sqrt{n}, then what?
After you have tried this, come back with more specific questions?
Edit: This is actually a very neat result and it works both ways, namely: A positive integer n greater than 1 is composite iff n has a divisor d satisfying 2<=d<sqrt{n}.
This tells us that whenever we want to check whether an integer n is composite or a prime, all we have to do is check whether it has a divisor up to sqrt{n}. Of course, if n is too large it is not that convenient, but for small n it works.
Ex: let n=65. Then, we have a rough idea of what sqrt{65} equals, namely >8. So all we need to do is see whether 2, 3,..., 8 divide 65.
