To make the question as close to the set theory as possible, let us restrict our attention to the case of a vector space V over the field F_2 of two elements. If my memory does not betray me, it is an old result by Paul Eklof that existence of a basis in an arbitrary vector space over F_2 is equivalent to the Axiom of Choice (it is easy in one direction: the Axiom of Choice, in the form of the Zorn Lemma, implies the existence of a basis). So, let us accept it, and let B be a basis in V, which means that every vector v in V is defined by its support in B, that is, by the (finite) set of elements in B which sum up to v. Therefore V has the same cardinality as the set of finite subsets of B; if B is infinite, than it is easy to prove that the set of all finite subsets of B has the same cardinality as B, hence V has the same cardinality as B.

Now let us look at the dual space V*, that is, the set of all linear functionals from V to F_2. Each such functional is uniquely determined by its support in B, that is, by the set of basis vectors where it takes value 1. Therefore V* is in one-to-one correspondence with the set 2^B of all subsets in B. But it is a classical result by Cantor that 2^B has larger cardinality than B and hence V* has larger cardinality than V. Of course, the cardinality of the bidual V** is even larger.

The case of an arbitrary field F can be handled in a similar way, but we will need to deal with the cardinality of the set F^B.

Is this intuitive? Well, it is intuitive for me because it was the first thought that crossed my mind. But I am not a set theorists and I have no idea whether the same can be proven without the Axiom of Choice. Also, the term “infinite dimensional vector space is ambigous: does it mean “a vector space without a finite basis” or “a vector space with an infinite basis”? And can anything that intimately involves the Axiom of Choice be called intuitive?