Let $V$ be any vector space over a field $k$. Let $\{e_i\}_{i \in I}$ be a basis for $V$. For each $i \in I$, there is a unique linear functional $f_i: V \rightarrow k$ such that $f_i(e_j) = \delta_{ij}$: that is, $f_i(e_i) = 1$ and for every other basis element $e_j$, $f_i(e_j) = 0$.
CLAIM: The set $\{f_i\}_{i \in I}$ is linearly independent in $V^{\vee}$, and thus $\dim V \leq \dim V^{\vee}$. It is a basis if and only if $V$ is finite-dimensional (if and only if $V^{\vee}$ is finite-dimensional).
The linear independence is easy: if $a_1 f_{i_1} + \ldots + a_n f_{i_n} = 0$, then just by evaluating at $e_{i_1},\ldots,e_{i_n}$ we find $a_1 = \ldots = a_n = 0$.
In the finite-dimensional case -- say $I = \{1,\ldots,n\}$ -- we may write any linear $g: V \rightarrow k$ as
$g = g(e_1) f_1 + \ldots + g(e_n) f_n$,
which shows that $f_1,\ldots, f_n$ is a basis for $V^{\vee}$ (and implies $\dim V = \dim V^{\vee}$).
However, in the infinite-dimensional case the $\{f_i\}_{i \in I}$ do not form a basis..essentially because in an abstract vector space all our sums must be finite sums rather than infinite sums! Indeed the subspace spanned by the $f_i$'s is precisely the set of linear functionals which are zero at all but finitely many basis elements $e_i$, whereas to give a linear functional $f$ on $V$ the values $f(e_i)$ can be absolutely arbitrary. Concretely, the functional $f$ with $f(e_i) = 1$ for all $i \in I$ does not lie in the span of the $f_i$'s.
(Remark: In fact whenever $\dim V$ is infinite, we have $\dim V^{\vee} > \dim V$. That is, not only is $\{f_i\}_{i \in I}$ not a basis, there is no basis for the dual space of cardinality equal to that of $I$. This is actually not so easy to prove, and it is not needed to answer the question.)
Now we come back to the canonical map $I: V \rightarrow V^{\vee \vee}$ given by
$I(x): f \mapsto f(x)$.
CLAIM: a) $I$ is always injective.
b) $I$ is surjective if and only if $V$ is finite-dimensional.
To prove a), let $x$ be a nonzero element of $V$ and choose a basis $\{e_i\}_{i \in I}$ for $V$ in which $x$ is one of the basis elements, say $x = e_1$. Then $f_1$ is a linear functional which does not vanish at $x$, so $I(x)$ is a nonzero element of $V^{\vee \vee}$.
To prove b) we first use the fact that if $V$ is finite-dimensional, $\dim V = \dim V^{\vee} = \dim V^{\vee \vee}$. Thus $I: V \rightarrow V^{\vee \vee}$ is an injection between two vector spaces of the same finite dimension, so it must be an isomorphism.
Finally, if $I$ is infinite-dimensional, then one can see by choosing bases, dual sets and dual dual sets as above that $I$ is not surjective. (A good first step here is to confirm that in the finite-dimensional case, if we choose a basis $e_1,\ldots,e_n$ for $V$, a dual base $f_1,\ldots,f_n$ for $V^{\vee}$ and then a dual dual base $g_1,\ldots,g_n$ for $V^{\vee \vee}$, then the map $I$ is precisely the one which maps $e_i$ to $g_i$ for all $i$.) I can supply more details upon request. Note also that if we are willing to make use of the above parenthetical fact that $\dim V^{\vee} > \dim V$ when $\dim V$ is infinite, then we see that $\dim V^{\vee \vee} > \dim V^{\vee} > \dim V$, and thus not only is $I: V \rightarrow V^{\vee \vee}$ not an isomorphism, but moreover there is no isomorphism of vector spaces from $V$ to $V^{\vee \vee}$. Again though, this lies significantly deeper.
Added: Let me say a bit about the more ambitious approach of showing $\operatorname{dim}_k V^{\vee} > \operatorname{dim}_k V$ for any infinite-dimensional vector space $V$. Let $I$ be a basis for $V$, so $V \cong \bigoplus_{I} k$. As mentioned above, to give a linear functional on $V$ it is necessary and sufficient to assign to each basis element an arbitrary value in $k$, whence an isomorphism $V^{\vee} \cong k^I = \prod_{I} k$. Thus dualization replaces a direct sum over $I$ with a direct product over $I$. When $I$ is finite there is no difference, so we recover $V \cong V^{\vee}$. However, when $I$ is infinite I claim that
$ \operatorname{dim}_k V^{\vee} = \operatorname{dim}_k k^I = \# k^{\# I} \geq 2^{\# I} > \# I = \operatorname{dim}_k V.$
I know of almost no standard texts which include a proof of this result, and indeed some cleverness / real ideas seem to be required (unlike the above discussion of the non-surjectivity of $I$ in the infinite-dimensional case which is, while somewhat lengthy to write out in detail, really very straightforward). However, by coincidence I just found on the web a very nice proof of this result which deduces it from Dedekind's Linear Independence of Characters. Please see this note of France Dacar.