Slick proof?: A vector space has the same dimension as its dual if and only if it is finite dimensional

Here is a simple proof I thought, tell me if anything is wrong.

First claim. Let $k$ be a field, $V$ a vector space of dimension at least the cardinality of $k$ and infinite. Then $\operatorname{dim}V^{*} >\operatorname{dim}V$.

Indeed let $E$ be a basis for $V$. Elements of V* correspond bijectively to functions from $E$ to $k$, while elements of $V$ correspond to such functions with finite support. So the cardinality of $V^{*}$ is $k^E$, while that of $V$ is, if I'm not wrong, equal to that of $E$ (in this first step I am assuming $\operatorname{card} k \le \operatorname{card} E$).

Indeed $V$ is a union parametrized by $\mathbb{N}$ of sets of cardinality equal to $E$. In particular $\operatorname{card} V < \operatorname{card} V^{*}$, so the same inequality holds for the dimensions.

Second claim. Let $h \subset k$ two fields. If the thesis holds for vector spaces on $h$, then it holds for vector spaces on $k$.

Indeed let $V$ be a vector space over $k$, $E$ a basis. Functions with finite support from $E$ to $h$ form a vector space $W$ over $h$ such that $V$ is isomorphic to the extension of $W$, i.e. to $W\otimes_h k$. Every functional from $W$ to $h$ extends to a functional from $V$ to $k$, hence

$$\operatorname{dim}_k V = \operatorname{dim}_h W < \operatorname{dim}_h W^* \leq \operatorname{dim}_k V^*.$$

Putting the two claims together and using the fact that every field contains a field at most denumerable yields the thesis.

It is clearly enough to show that an infinite dimensional vector space $V$ has smaller dimension that its dual $V^*$.

Let $B$ be a basis of $V$, let $\mathcal P(B)$ be the set of its subsets, and for each $A\in\mathcal P(B)$ let $\chi_A\in V^*$ be the unique functional on $V$ such that the restriction $\chi_A|_B$ is the characteristic function of $A$. This gives us a map $\chi:A\in\mathcal P(B)\mapsto\chi_A\in V^*$.

Now a complete infinite boolean algebra $\mathcal B$ contains an independent subset $X$ such that $|X|=|\mathcal B|$---here, that $X$ be independent means that whenever $n,m\geq0$ and $x_1,\dots,x_n,y_1,\dots,y_m\in X$ we have $x_1\cdots x_n\overline y_1\cdots\overline y_n\neq0$. (This is true in this generality according to [Balcar, B.; Franěk, F. Independent families in complete Boolean algebras. Trans. Amer. Math. Soc. 274 (1982), no. 2, 607--618. MR0675069], but when $\mathcal B=\mathcal P(Z)$ is the algebra of subsets of an infinite set $Z$, this is a classical theorem of [Fichtenholz, G. M; Kantorovich L. V. Sur les opérations linéaires dans l'espace des fonctions bornées. Studia Math. 5 (1934) 69--98.] and [Hausdorff, F. Über zwei Sätze von G. Fichtenholz und L. Kantorovich. Studia Math. 6 (1936) 18--19])

If $X$ is such an independent subset of $\mathcal P(B)$ (which is a complete infinite boolean algebra), then $\chi(X)$ is a linearly independent subset of $V^*$, as one can easily check. It follows that the dimension of $V^*$ is at least $|X|=|\mathcal P(B)|$, which is strictly larger than $|B|$.

Later: The proof of the existence of an independent subset is not hard; it is given, for example, in this notes by J. D. Monk as Theorem 8.9. In any case, I think this proof is pretty because it captures precisely the intuition (or, rather, my intuition) of why this is true. I have not seen the paper by Fichtenhold and Kantorovich (I'd love to get a copy!) but judging from its title one sees that they were doing similar things...

I know a fairly elementary proof in the case when the field is countable.

First, you prove that $Hom(\bigoplus_{i\in I}A_{i},B)\cong \prod_{i\in I}Hom(A_{i},B)$, where all terms are $R$-modules. (This should be fairly intuitive. A homomorphism from a direct sum is determined by its actions on each piece individually.)

Second, specialize $A_{i}$ and $B$ to equal your field. So the direct product is over a bunch of pieces (all isomorphic to your field).

Third, use the standard cardinality argument to show that a direct product of $I$ non-empty pieces has cardinality strictly greater than $I$.

This argument doesn't quite work when your field has large cardinality, but I still think it is nice. (Basically, this is thinking about the first part of Andrea's proof a little differently.)