Question :
Assume that $V$ is a finite-dimensional vector space.
Prove that each basis of $V^*$ is dual of a basis of $V$.
My try :
I defined an isomorphism this way :
$i : V \to V^{**}$
For each $v \in V$, $i$ takes $v$ and gives $i(v)$.
What is $i(v)$? Its a linear-transformation.
$i(v):V^* \to F$
$(i(v))(f)=f(v)$
I proved that this transformation is linear and one-to-one. Then i concluded that its an isomorphism. ( The details are not important cause i've done that and its not my problem. )
To prove the theorem, Take an arbitrary basis of $V^*$ like $C$. Find the dual of $C$ ( call it $C^*$ ) which is in $V^{**}$. Using the isomorphism $i$, Consider the inverted image of the members of $C^*$ in $V$. This set is a basis of $V$. Call it $B$.
Now, I need to prove that $B^*=C$. That's where i'm stuck.
Note 1 : There is a similar question. But its closed and i don't wan't that methods for proving this theorem. I want to complete my proof.
Your idea is good! In order not to confuse indices, let me call $\omega\colon V\to V^{**}$ the isomorphism.
Consider $\{\xi_1,\dots,\xi_n\}$ a basis of $V^*$ and its dual basis $\{\xi_1^*,\dots,\xi_n^*\}$ in $V^{**}$. This has the property that $$ \xi_i^*(\xi_j)=\delta_{ij} \qquad(1\le i\le n, 1\le j\le n) $$ (Kronecker delta) and is uniquely determined by this property. Since we have an isomorphism $\omega\colon V\to V^{**}$, we can define $v_i=\omega^{-1}(\xi_i^*)$ and the set $\{v_1,\dots,v_n\}$ is automatically a basis for $V$.
Now, by definition of the isomorphism $\omega$, $$ \xi_i(v_j)=\omega(v_j)(\xi_i)=\xi_j^*(\xi_i)=\delta_{ji}=\delta_{ij} $$ so indeed $\{\xi_1,\dots,\xi_n\}$ is the dual basis for $\{v_1,\dots,v_n\}$.