When is a subspace of $V^*$ dense, where $V$ is a Banach space? Let $V$ be a Banach space. Suppose $\alpha_i$ is a collection of bounded linear functionals $V\to \mathbb{C}$ such that:
$$ \bigcap_i \ker(\alpha_i) = 0. $$
Does this imply the $\alpha_i$ span a dense subset of the dual $V'$? 
I'm motivated by the following special case: $V = C([0, 1])$, the space of all continuous $[0, 1] \to \mathbb{C}$. Now for $x\in [0, 1]$, the "evaluation function" $e_x : V \to \mathbb{C}$ simply takes $v \mapsto v(x)$. I'd like to prove that the set of evaluation functions spans a dense subset of $V'$ (the usual way is to use Riesz representation theorem to describe $V'$ as the space of complex measures on $[0, 1]$, but I'm trying not to use that). Note that in our case, we clearly do have $\bigcap_i \ker \alpha_i = 0$.
 A: Not in general: Take $V = \ell^1$, and for each $i\in \mathbb{N}$, let $\alpha_i$ denote the "evaluation" map
$$
\alpha_i ((x_n)) := x_i
$$
Then clearly
$$
\bigcap_i \ker(\alpha_i) = 0
$$
However, consider the dual space pairing
$$
\ell^{\infty} \to (\ell^1)^{\ast}
$$
then the $\alpha_i$ correspond to the elements
$$
e_i = (0,0,0, \ldots, 0, 1,0,\ldots) \in \ell^{\infty}
$$
In particular,
$$
\text{span}(\alpha_i) \subset c_0 \neq \ell^{\infty}
$$
However, what you say is true if $V$ is reflexive : Since
$$
\bigcap_i \ker(\alpha_i) = 0
$$
it follows from reflexivity that for any $T \in V^{\ast\ast}$,
$$
T(\alpha_i) = 0\quad\forall i \Rightarrow T \equiv 0
$$
It is now a consequence of Hahn-Banach that this implies
$$
\overline{\text{span}(\alpha_i)} = V^{\ast}
$$
(else you could construct a non-zero linear functional on $V^{\ast}$ that annihilates all the $\alpha_i$)
A: This is not in fact true in the case that you are interested in.  For instance, consider the functional $I:C([0,1])\to\mathbb{C}$ given by $I(f)=\int f d\mu$, where $\mu$ is Lebesgue measure.  It is easy to see that any finite linear combination of evaluation functionals has distance $\geq 1$ from $I$ (because you can find an element of $C([0,1])$ of norm $1$ which vanishes at the finitely many points you're evaluating at but nevertheless has integral $1-\epsilon$).
