Let $V$ be a (possibly infinite-dimensional) vector space. Let $U \subseteq V^*$ be a finite-dimensional vector subspace. Define $W = \{v \in V: f(v)=0 \text{ for all } f \in U \}$. Does it then follow that $V/W \cong U^*$?
Here is my attempt at a proof:
Consider the map $\psi: V/W \to U^*$ given by $v+W \mapsto (f \mapsto f(v))$. This is a well-defined linear map and is injective just by the definition of $W$. We now want to show that it is surjective. Let $e: V \to V^{**}$ be the natural map. I claim that $U^* \subseteq e(V)$. If this were true I claim that $q \circ e^{-1}: U^* \to V/W$ is inverse to $\psi$.
I am now sure how to prove the key claim that $U^* \subseteq e(V)$ (nor even if it is true). Perhaps there is a better way to prove this proposition?
Your claim $U^*\subset e(V)$ may be proved as follows.
Let $(f_1,\dots,f_n)$ be a basis of $U.$
Since $f_1$ is not a linear combination of $f_2,\dots,f_n,$ its kernel does not contain the intersection of theirs, i.e. $$\exists v_1\in V\quad f_1(v_1)=1\quad\text{and}\quad f_2(v_1)=\dots=f_n(v_1)=0.$$ Similarily, $\exists v_2,\dots,v_n\in V$ $$\forall i,j\quad f_i(v_j)=\delta_{i,j}.$$ Then, $$\forall\varphi\in U^*\quad\varphi=e\left(\sum_{i=1}^n\varphi(f_i)v_i\right).$$