I am given an inclusion of vector spaces $L \subseteq V$ and I know that $V^{\vee \vee} = W^\vee$ for some vector space $W$. There are no assumptions of finite dimensionality. Through the inclusion $$ L \subseteq V \subseteq V^{\vee \vee} = W^\vee$$ I can define $$L_0:= \{w \in W \mid \forall \ \ell \in L, \ \ell(w) = 0 \}.$$ Under the assumption that $L_0$ is finite dimensional, I would like to show that $V/L \cong L_0^\vee$ canonically. Any hints?
Remark: It must be crucial that $L_0$ is finite dimensional. For instance, for $L=0$ and $V$ infinite dimensional we would have $$V/L=V \subsetneq V^{\vee\vee} = L_0^\vee = W^\vee.$$
Attempt 1: I have been trying to use the fact that for every inclusion $A \subseteq B$ of vector spaces, we always have
$$ A^\vee \cong B^\vee / A^{\bot_B}, \quad A^{\bot_B} \cong (B/A)^\vee,$$
where $A^{\bot_B}:=\{ \phi \in B^\vee \mid \forall \ a \in A, \ \phi(a) = 0\}$, so in my case I could write something like $$ L_0^\vee \cong W^\vee / L_0^{\bot_W} = V^{\vee\vee} / L_0^{\bot_W},$$ but I am not sure whether this helps at all.
Attempt 2: We have $$ L_0 \cong L^{\bot_{W^\vee}} \cap W \subseteq W^{\vee \vee},$$ but also this line of thought did not bring me anywhere.