"A gentleman never chooses a basis." Around these parts, the aphorism "A gentleman never chooses a basis," has become popular.

Question. Is there a gentlemanly way to prove that the natural map from $V$ to $V^{**}$ is surjective if $V$ is finite-dimensional?

As in life, the exact standards for gentlemanliness are a bit vague. Some arguments seem to be implicitly picking a basis. I'm hoping there's an argument which is unambiguously gentlemanly.
 A: Following up on Qiaochu's query, one way of distinguishing a finite-dimensional $V$ from an infinite one is that there exists a space $W$ together with maps $e: W \otimes V \to k$, $f: k \to V \otimes W$ making the usual triangular equations hold. The data $(W, e, f)$ is uniquely determined up to canonical isomorphism, namely $W$ is canonically isomorphic to the dual of $V$; the $e$ is of course the evaluation pairing. (While it is hard to write down an explicit formula for $f: k \to V \otimes V^*$ without referring to a basis, it is nevertheless independent of basis: is the same map no matter which basis you pick, and thus canonical.) By swapping $V$ and $W$ using the symmetry of the tensor, there are maps $V \otimes W \to k$, $k \to W \otimes V$ which exhibit $V$ as the dual of $W$, hence $V$ is canonically isomorphic to the dual of its dual. 
Just to be a tiny bit more explicit, the inverse to the double dual embedding $V \to V^{**}$ would be given by 
$$V^{\ast\ast} \to V \otimes V^* \otimes V^{\ast\ast} \to V$$ 
where the description of the maps uses the data above. 
A: Over real or complex (or other similar) field, where we know that for a finite-dimensional vector space all reasonable vector-space topologies coincide... V is dense in V** in the weak topology, hence in all topologies, but the (unique) topology is also complete, so V = V** (I think this works and avoids choosing a basis.  Of course you would have to prove those other facts also without choosing a basis.)
A: Perhaps it would be most appropriate to answer your question with another question: how do you distinguish a finite-dimensional vector space from an infinite-dimensional one without talking about bases?
A: At the price of being too categorical for the question, one can follow up Todd's answer as follows.
Consider any closed symmetric monoidal category $\mathcal{V}$ with product $\otimes$ and unit object $k$, such as vector spaces over a field $k$. Let $V$ be an object of $\mathcal{V}$ and let $DV = Hom(V,k)$.  Just from formal properties of $\mathcal{V}$, there are canonical maps $\iota\colon k\to Hom(V,V)$ and $\nu\colon DV\otimes V\to Hom(V,V)$, which are the usual things for vector spaces.  Say that $V$ is dualizable if there is a map  $\eta\colon k\to V\otimes DV$ such that $\nu \circ \gamma \circ \eta = \iota$, where $\gamma$ is the commutativity isomorphism.  Formal arguments show that $\nu$ is then an isomorphism and if $\epsilon\colon DV\otimes V \to k$ is the evaluation map (there formally), then, with $W=DV$, $\eta$ and $\epsilon$ satisfy the conditions Todd stated for $e$ and $f$.  This is general enough that it can't have anything to do
with bases.  But restricting to vector spaces, we can choose a finite set of elements $f_i\in DV$
and $e_i\in V$ such that $\nu(\sum f_i\otimes e_i) = id$.  Then it is formal that $\{e_i\}$ is 
a basis for $V$ with dual basis $\{f_i\}$.  This proves that $V$ is finite dimensional, and the 
converse is clear as in Todd's answer. There is a result in Cartan-Eilenberg called the dual basis theorem that essentially points out that the precisely analogous characterization holds for finitely generated projective modules over a commutative ring $k$, with the same proof.
Still in a general symmetric monoidal category, if $V$ is dualizable, then a formal argument also shows that the canonical map $V \to V^{**}$ (again defined formally) is an isomorphism.  Also, in answer to Peter Samuelson, while the name ``dual basis theorem'' dates from long before my time, it does have some justification.  When  $\mathcal{V}$ is modules
over a commutative ring $k$, if one takes a dualizable $V$ and constructs the free module $F$ on basis $\{d_i\}$ in 1-1 correspondence with the $e_i$ in my previous post, then $\alpha(v) = \sum f_i(v) d_i$
specifies a monomorphism $\alpha\colon V\to F$ such that $\pi\alpha = id$, where $\pi(d_i) = e_i$.  This completes the proof that dualizable implies finitely generated projective, with a relevant basis in plain sight.
A: there is a canonical map $ev:V \to V^{**}$ defined by $ev(v)(\phi) = \phi(v)$.  to check that it is an isomorphism in the finite dimensional setting you can just check that it is injective and this is evident from the definition.
A: Some kind of solution proposal:
Let V be a n-dimensional vector space over a field (or a free R-module, where R is a commutative unital ring).
There is a morphism V tensor V* to End(V), which sends each v tensor lambda to the endomorphism of V that sends each w to lambda(w)v. It is an epimorphism since it's image are all finite rank endomorphisms, so it's surjective. It is a monomorphism as you can check by calculation. So this is an isomorphism.
We can calculate the dimensions: dim(V tensor V* ) = dim(End(V)), where dim(V tensor V* ) = dimV * dimV* and dim(End(V))=(dimV)^2. So the result is n * dimV* = n^2 and we get dimV* = n = dimV.
Now notice that every short exact sequence in our category splits.
That implies for every monomorphism V to W, that W is isomorphic to a direct sum of V and W/V and therefore we have a dimension formula dimV + dim(V/W) = dim W. We get the result that every monomorphism from V to W with dimW=dimV is an isomorphism.
Look at the linear map ev : V to V**, which sends v to ev_v : (lambda mapsto lambda(v)), the evaluation-at-v-map. Now we make an induction: for dimV=0, the map ev is trivially an isomorphism. For dimV=n, the kernel of ev is a subspace, so we have V = ker(ev) + W with some complement W and either ker(ev)=V or ker(ev)=0 or the two subspaces have strictly smaller dimension. That would mean, by induction hypothesis, that their evaluation map, which is the restriction of the evaluation map of V, has no kernel and so we get ker(ev)=0. The case ker(ev)=V remains, where we get that V*=0 which contradicts n=dimV=dimV*.
Now ev is a monomorphism and dim(V** )=dim(V* )=dim(V), therefore ev is an isomorphism.
One can check easily that this is "functorial", that is: we have a natural transformation from the identity functor to the bidual functor.
One could object that I have chosen an arbitrary flag, when I take the complement of the kernel in the induction step... but I guess without that you wouldn't use the "free" property of the modules in question, and for non-free modules there are counter-examples.
If I did something wrong, please tell me.
