Every subspace of the dual of a finite-dimensional vector space is an annihilator

For a subspace $T\subseteq V'$ define $T^0 := \{v\in V\,|\, \varphi(v) = 0\text{ for all $\varphi\in T$}\}$ (which is the same as your definition under the identification $V = V''$). In your case, we have $\Gamma^0 = S$.

Then we have $\dim V = \dim U + \dim U^0$ for each subspace $U\subseteq V$: Take a basis $u_1,\dotsc,u_d$ of $U$ and complete it to a basis $u_1,\dotsc,u_n$ of $V$. Denoting by $u'_1,\dotsc,u'_n$ the dual basis, it is easy to see that $U^0$ is spanned by $u'_{d+1},\dotsc,u'_n$, which proves the claim.

Likewise, we have $\dim V' = \dim T + \dim T^0$ for any subspace $T\subseteq V'$.

Now, it follows that $T^{00} = T$ for $T\subseteq V'$: $T\subseteq T^{00}$ is clear and both have the same dimension, because of $$\dim T = \dim V' - \dim T^0 = \dim V - \dim T^0 = \dim T^{00}.$$

With this, we conclude $\Gamma = \Gamma^{00} = S^0$ (using $\Gamma^0 = S$ for the last equality).

I really didn't understand the user218931's solution. But, I think I solved this problem in another way:

Let $S = \{v \in V: \varphi(v)=0\text{ for every }\varphi\in\Gamma \}$ and let $F$ be the field. Let $\phi_1, \phi_2, \cdots, \phi_m$ be a basis for $\Gamma$.

By the definition of annihilator, $\Gamma \subset S^{0}$. Now, to prove that $ S^{0} \subset \Gamma$, first it is necessary prove that S is a subspace of V:

• $\varphi(0)=0 \Rightarrow 0 \in S$
• If $v_1 \in S$ and $v_2 \in S$, then for every $\varphi \in \Gamma$ we have $\varphi(v_1 + v_2) = \varphi(v_1) + \varphi(v_2) =0$. So, $v_1+v_2 \in S$.
• if $v \in S$ and $\lambda \in F$, then for every $\varphi \in \Gamma$ we have $\varphi(\lambda v)= \lambda\varphi(v)=0$. Thus, $\lambda v \in S$.

Therefore, $S$ is a subspace of $V$. Therefore we can use:

$$\dim S + \dim S^{0}= \dim V$$

Now, notice that the definition of $S$ is equivalent to $S = null \space \phi_1 \space \cap null \space \phi_2 \space \cap \cdots \cap \space null \space \phi_m$, because $v \in S$ if and only if $\phi_i(v)=0 $ for all $i \in \{1,2,..., m\}$ . Besides, it is known that

$$\dim(null \space \phi_1 \space \cap null \space \phi_2 \space \cap \cdots \cap \space null \space \phi_m) = \dim V - m $$

Thus, $\dim S^{0} = m =\dim \Gamma$. Since the set $\Gamma \subset S^0$, then $\Gamma= S^0$.