Definition: Let $(V,F,+_V,\cdot_V)$ and $(F,F,+_F,\cdot_F)$ be vector space over field $F$. Then $V^{**}$ $=(V^*)^*$ $=L(V^*,F)$ $=L(L(V,F),F)$ is called double dual space of $V$.
Let $V$ be a finite-dimensional vector space over the field $F$. Each basis for $V^*$ is the dual of some basis for $V$.
My attempt: Since $\mathrm{dim}(V)=n\in \Bbb{N}$, we have $\mathrm{dim}(V^*)$ $= \mathrm{dim}(V)$ $=n$. Let $B^*=\{f_1,…,f_n\}$ be a basis of $V^*$. Let $B^{**}=\{L_1,…,L_n\}$ be dual basis of $B^*$. So $L_i(f_j)=\delta_{ij}$, $\forall i,j\in J_n$. By theorem 17 corollary 1 section 3.6, $\forall i \in J_n$, $\exists !\alpha_i\in V$ such that $\phi(\alpha_i)=L_{\alpha_i}=L_i$. We claim $B=\{\alpha_1,…,\alpha_n\}$ is basis of $V$. Let $\alpha \in V$. Then $\phi(\alpha)=L_\alpha \in V^{**}$. Since $\mathrm{span}(B^{**})=V^{**}$, we have $L_\alpha$ $=\sum_{i\in J_n}a_i\cdot L_i$, for some $a_i\in F$. By theorem 17 section 3.6, $\phi:V\to V^{**}$ is isomorphism. By theorem 7 section 3.2, $\phi^{-1}$ is linear map. So $\phi^{-1}(L_\alpha)$ $=\phi^{-1}(\sum_{i\in J_n}a_i\cdot L_i)$ $=\alpha$ $= \sum_{i\in J_n}a_i\cdot \phi^{-1}(L_i)$ $= \sum_{i\in J_n}a_i\cdot \alpha_i$. Thus $\alpha \in \mathrm{span}(\{\alpha_1,…\alpha_n\})$. Hence $V=\mathrm{span}(\{\alpha_1,…,\alpha_n\})$. If $\sum_{i\in J_n}c_i\cdot \alpha_i=0$, for some $c_i\in F$. So $\phi (0)$ $=\phi(\sum_{i\in J_n}c_i\cdot \alpha_i)$ $= \sum_{i\in J_n}c_i\cdot \phi(\alpha_i)$ $= \sum_{i\in J_n}c_i\cdot L_{\alpha_i}$ $= \sum_{i\in J_n}c_i\cdot L_i$ $=0$. Since $B^{**}$ is independent, we have $c_i=0_F$, $\forall i\in J_n$. Thus $\{\alpha_1,…\alpha_n\}$ is independent. Hence $B=\{\alpha_1,…,\alpha_n\}$ is basis of $V$. We claim $B^*=\{f_1,…,f_n\}$ is dual basis of $B=\{\alpha_1,…,\alpha_n\}$. Let $i,j\in J_n$. Since $L_{\alpha_j}=L_j$, we have $L_j(f)$ $=L_{\alpha_j}(f)$ $=f(\alpha_j)$, $\forall f\in V^*$. In particular, $L_j(f_i)$ $=L_{\alpha_j}(f_i)$ $=f_i(\alpha_j)$. Since $L_j(f_i)=\delta_{ij}$, we have $f_i(\alpha_j)=\delta_{ij}$. Thus $f_i(\alpha_j)=\delta_{ij}$, $\forall i,j\in J_n$. Hence $B^*=\{f_1,…,f_n\}$ is dual basis of $B=\{\alpha_1,…,\alpha_n\}$. Is my proof correct?