linear algebra - Concerning the existence of a basis corresponding to the dual basis $phi_1,phi_2,...,phi_nin V'$

Is the following Proof Correct?

Theorem. Given that $V$ is finite dimensional and $\phi_1,\phi_2,...,\phi_n$ is a basis for $V' = \mathcal{L}(V,\mathbf{F})$ i.e. the dual of $V$. Show that there exists a basis of $V$ whose dual basis is $\phi_1,\phi_2,...,\phi_n$.

Proof. Let $I_n = \{i\in\mathbf{Z^+}|i\leq n\}$. Since $\phi_1,\phi_2,...,\phi_n$ is a basis for $V'$ we can affirm that $\forall j\in I_n$ $\phi_j$ is not the zero linear functional consequently there exist a list of vectors $v_1,v_2,...,v_n$ such that $\forall j\in I_{n}(\phi_j(v_j)\neq 0)$.

Since $\dim V = \dim V'$ in order to show that $v_1,v_2,...,v_n$ is a basis it is sufficient for us to show that $v_1,v_2,...,v_n$ is linearly independent.

Assume that for some $c_1,c_2,...,c_n\in\mathbf{F}$ $$c_1v_1+c_2v_2+\cdot\cdot\cdot+c_nv_n = 0\tag{1}$$
and that $i\in I_n$, applying $\phi_{i}$ to $(1)$ yields the equation $0 = c_i\phi(v_i)$ since $\phi_i(v_i)\neq 0$ it follows that $c_i = 0$ and since our choice of $i$ was arbitrary it follows that $c_1 = c_2 = ... = c_n = 0$ .

Answer

No, the proof is not correct.
It is not enough that $\phi_i(v_i)=1$; you also need information about $\phi_i(v_j)$ when $i\neq j$.
In your computation for linear independence you assumed that $\phi_i(v_j)=0$ when $i\neq j$, but this does not follow from your choice of $v_i$s.

If you can find a basis $w_1,\dots,w_n$ of $V$ so that $\phi_i(w_j)=\delta_{ij}$, then your proof works, but you need to elaborate on why such a basis exists.
For example, if $w_i$s are such a good basis, in your proof nothing prevents you from choosing
$$v_1=v_2=\dots=v_n=w_1+w_2+\dots+w_n.$$
Now all the $v_i$s are equal but $\phi_i(v_i)=1$.
The problem is that $\phi_i(v_j)=1$ for all $i,j\in I_n$, and no terms vanish in the sum.