Let S be the subset of natural numbers that are not interesting. Suppose by way of contradiction that S is inhabited. Then by the well ordering principle of natural numbers, there is a least such element, s in S. In virtue of being the least non interesting number, s is in fact interesting. Hence s is not in S. Since s is in S and not in S, we have derived a contradiction. Therefore our assumption that S is inhabited must be false. Thus S is empty and there are no non interesting numbers.
My favorite version of this proof:
Let S be the subset of natural numbers that are not interesting. Suppose by way of contradiction that S is inhabited. Then by the well ordering principle of natural numbers, there is a least such element, s in S. In virtue of being the least non interesting number, s is in fact interesting. Hence s is not in S. Since s is in S and not in S, we have derived a contradiction. Therefore our assumption that S is inhabited must be false. Thus S is empty and there are no non interesting numbers.