But the simplest way would be to use the fundamental theorem of arithmetic (that every integer has a unique prime factorization). For any N, if sqrt(N) is rational, you can write that as
N=A^2/B^2
and therefore
B^2 N = A^2
and it's not hard to get from that + the fundamental theorem to the conclusion that N is a square.
It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC).
Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?