M 307 Proof of the Well Ordering Principle of the Natural Numbers

Below is my proof of the Well Ordering Principle.  It is done via Induction and Contradiction.  Again, the box did not survive the file conversion.  Every time you see a box, there should be a large N, indicating the set of Natural Numbers (or positive Integers--counting numbers beginning with 1).

I really like this proof.  I find it relatively easy to understand, and one that I feel I could explain to high school students fairly simply.  It pretty much says every subset of the counting numbers must have a minimum, or smallest element.  The proof proceeds by contradiction by assuming there exists a subset of the Natural numbers which does not contain a minimum.  This subset must have a lower bound, or number outside of the set that is smaller than every number in the set.  The proof then proceeds by induction.  Is every number in the subset larger than 1?  Yes, since the subset does not contain a minimum and is a subset of the Natural numbers, every element in the subset must be larger than the lower bound 1.  Then we assume that our inductive step holds, and Q(2), Q(3), ..., Q(n) holds, which means that the numbers 1, 2, 3, ..., n are all lower bounds of our subset.  We must then show that Q(n+1) holds.  But this means that every natural number is a lower bound of our subset.  This is a contradiction, because our set is a subset of the natural numbers, and thus must contain natural numbers.  I think this makes fairly simple sense.  The big problem in understanding I see in high school students would be accepting the inductive step.  I still have some trouble with this--assuming that the assertion holds in order to prove the assertion seems logically incorrect.