How to prove that $(S^{0})^{0} = \operatorname{span}(\psi(S))$, where $\psi: V \to V^{**}$ is the natural isomorphism 
Given that $V$ is a finite dimensional vector space. The annihilator $S^0$ of $S$ is the set 
  $$S^0 = \{f \in V^* :\, (\forall  x \in S) \, f(x)=0\}.$$ 
  If $W$ is a subspace of V and $x \notin W$, prove that there exists $f \in W^0$ such that $f(x) \neq 0$. Prove that $(S^0)^0 = \operatorname{span} (\psi(S))$, where $\psi: V \to V^{**}$ is the natural isomorphism.

How am I supposed to prove $(S^{0})^{0} = \operatorname{span}(\psi(S))$?
 A: 
Lemma. Let $V$ be a finite-dimensional vector space, and $W$ be a subspace of $V$. If $x \notin W$, then there exists some $f \in W^0$ such that $f(x) \neq 0$.

Proof. Let $\gamma = \{w_1,\dots,w_n\}$ be a basis for $W$. Since $x \notin W$,
$\gamma \cup \{x\} = \{w_1,\dots,w_n,w_{n+1}\}$ (where $w_{n+1} := x$) is a linearly independent subset that can be extended to a basis
$$\beta = \{w_1,\dots,w_n,w_{n+1},\dots,w_m\}$$
for $V$. Consider the dual basis $\beta^* = \{\phi_1,\dots,\phi_n,\phi_{n+1},\dots,\phi_m\}$ of $\beta$. By definition, $\phi_{n+1}(x) = 1$ and $\phi_{n+1}(w_i)=0$ for $i=1,\dots,n$. Of course, the latter implies that $\phi_{n+1} \in W^0$, so let $f := \phi_{n+1}$. $\blacksquare$

Problem. For any subset $S$ of $V$ we have that $(S^0)^0 = \operatorname{span} \psi(S)$, where $\psi : V \to V^{**}$ is the canonical isomorphism.

Proof. We will prove first that $\psi(S) \subseteq (S^0)^0$. Suppose then that $x$ is an element of $\psi(S)$. This means that there exists some $s \in S$ such that $x = \psi(s)$.
In order to prove that $x$ is in $(S^0)^0$ we need to prove that $x(f) = 0$ for all $f \in S^0$, so, pick any $f \in S^0$. Now, observe that
$$x(f) = \psi(s)(f) = f(s)$$
and the latter is in fact equal to $0$ because any element of $S^0$ annihilate every element of $S$. This proves that $\psi(S) \subseteq (S^0)^0$, and it follows$^1$ that
$$\operatorname{span} \psi(S) \subseteq (S^0)^0.$$
To prove the other direction, suppose that $x$ is an element of $(S^0)^0$. Since $\psi$ is surjective, $x = \psi(v)$ for some $v \in V$. Because $x$ is in $(S^0)^0$ we have that
$$\forall f \in S^0 : 0 = x(f) = \psi(v)(f) = f(v). \tag{$\dagger$}$$
Now, in order to prove that $x$ is in $\operatorname{span} \psi(S)$, which is the same$^2$ as $\psi(\operatorname{span} S)$, we need to prove that $v \in \operatorname{span} S$ (since $x = \psi(v)$).
What happens if $v \notin \operatorname{span} S$? Well, by the lemma, there is some $f \in (\operatorname{span} S)^0$ such that $f(v) \neq 0$, but as $(\operatorname{span} S)^0 \subseteq S^0$, this contradicts $(\dagger)$.
Hence $v \in \operatorname{span} S$, and then $x \in \psi(\operatorname{span} S) = \operatorname{span}\psi(S)$. $\blacksquare$

$^1$ The set $\operatorname{span} \psi(S)$ is the smallest subspace of $V^{**}$ that contains $\psi(S)$, in the sense that any other subspace of $V^{**}$ containing $\psi(S)$ also contains $\operatorname{span} \psi(S)$.
$^2$ The inclusion $\operatorname{span} \psi(S) \subseteq \psi(\operatorname{span} S)$ follows from the preceding footnote (why?). The other inclusion is also easy. Take any $y \in \psi(\operatorname{span} S)$, which means that $y = \psi(w)$ for some $w$ that can be written as a linear combination of some elements of $S$. That $y \in \operatorname{span} \psi(S)$ follows from the linearity of $\psi$, can you see why?
