Let $V$ be a finite-dimensional vector space. Let $V^*$ be the dual space of $V$. Choose a basis $ \mathcal{B} = \{\textbf{e}_1,\dots,\textbf{e}_n\}$ for $V$.
I would like to prove that there exist $\theta_1,\dots,\theta_n \in V^*$ such that $$\textbf{v} = \theta_1 (\textbf{v}) \textbf{e}_1 + \dots + \theta_n (\textbf{v}) \textbf{e}_n$$ for some $\textbf{v} \in V$.
I am having trouble constructing such $\theta$'s, or finding a contradiction. This is a key stepping stone for proving that dual sets exist, as shown here. I do not see how this immediately follows from $\mathcal{B}$ being a basis of $V$.
You want the relation $$ \mathbf{v} = \theta_1 (\mathbf{v}) \mathbf{e}_1 + \dots + \theta_n (\mathbf{v}) \mathbf{e}_n $$ to hold for all $\mathbf{v}\in V$. In particular it should hold for $\mathbf{e}_1$, so $$ \mathbf{e}_1 = \theta_1 (\mathbf{e}_1) \mathbf{e}_1 + \dots + \theta_n (\mathbf{e}_1) \mathbf{e}_n $$ forces $$ \theta_1(\mathbf{e}_1)=1,\quad \theta_2(\mathbf{e}_1)=0,\quad \dots,\quad \theta_n(\mathbf{e}_1)=0,\quad $$ Similarly, you get that $$ \theta_i(\mathbf{e}_j)=\begin{cases} 1 & i=j \\[4px] 0 & i\ne j \end{cases} $$ Note that these relations are sufficient for defining linear maps $\theta_i\colon V\to F$, for $i=1,2,\dots,n$ (where $F$ is the base field), because we are prescribing their action on the elements of a basis.
Prove that, if $\mathbf{v}=\alpha_1\mathbf{e}_1+\alpha_2\mathbf{e}_2+\dots+\alpha_n\mathbf{e}_n$, then $$ \theta_i(\mathbf{v})=\alpha_i $$ and the rest should be easy.