Proof verification of the fact that if $W\subset V$, then $W^{00}$ is the subspace spanned by $W$. I've put together this proof and was wondering if some of you could verify and give me tips on where to improve:
Theorem: Let $W$ a subspace of $V$ over $F$, then $W^{00}=span(W)$.
Proof:$\quad$ Take $S\subset V$ spanned by $W$, then if $f(\alpha)=0, \quad \alpha\in S$, certainly  $f(\beta)=0, \quad \beta\in W$. Therefore $$S^0=W^0$$
Furthermore, if we let $\alpha_1,\alpha_2,...,\alpha_n$ the maximum number of linearly independent vectors in $W$, then $f(\alpha_i)=0, \quad i=1, 2, ...n$. Certainly any linear combination $\alpha$ of these vectors yields $f(\alpha)=0$.
Now we know $$S^{00}=S\qquad S^{00}=W^{00}$$ and therefore$$W^{00}=S$$Thus, $W^{00}$ is the span of $W$.
 A: You're apparently assuming $V$ is finite dimensional. However, the final part of your attempt is insufficient to be a proof.
If $S$ is a subset of $V$ and $W$ is the subspace spanned by $S$, then
$$
S^0=W^0
$$
where
$$
S^0=\{\xi\in V^*:\xi(x)=0,\text{ for all }x\in S\}=
\{\xi\in V^*:\ker\xi\subseteq S\}
$$
One inclusion is obvious, because $W\supseteq S$; the other one follows from linearity of elements in $V^*$, so $S\supseteq \ker\xi$ implies $W\supseteq\ker\xi$.
So it's not restrictive to assume $W$ is a subspace of $V$. Now we set
$$
W^{0\,0}=\{x\in V:\xi(x)=0,\text{ for all }\xi\in W^0\}
$$
and we wish to prove that $W=W^{0\,0}$. The inclusion $W\subseteq W^{0\,0}$ is obvious.
Let $\{v_1,\dots,v_k\}$ be a basis of $W$.
We need to show that, if $y\in V\setminus W$, then there is $\xi\in W^{0\,0}$ such that $\xi(y)\ne0$.
Hint: set $y=v_{k+1}$; then $\{v_1,\dots,v_k,v_{k+1}\}$ is linearly independent, so it can be extended to a basis $\{v_1,\dots,v_n\}$ of $V$; define $\xi\in V^*$ so that…
