Suppose that $(W,\Omega)$ is a symplectic vector space, $L \subset W$ is a Lagrangian subspace and $G \subset W$ is a coisotropic subspace ($G^\perp \subset G$) such that $L \cap G^\perp = \{0\}$. (Where for any $V \subset W$, $V^\perp := \{w \in W | \Omega(v,w) = 0, \forall v \in V\}$).
I would like to show that $(W',\Omega')$, $W' := G/G^\perp, \Omega'([v],[u]) := \Omega(v,u)$ is a symplectic vector space and that $L' := L \cap G$ is a Lagrangian subspace in $W'$.
I think can do this if $W$ is finite dimensional:
$\Omega'$ is a well-defined symplectic form because if $v' = v + v'', u' = u + u'' \in G$ where $v'', u'' \in G^\perp$ then $\Omega'([v'],[u']) = \Omega(v',u') = \Omega(v + v'',u + u'') = \Omega(v,u) = \Omega'([v],[u])$. $\Omega'$ is non-degenerate because if $v \in G$ is such that $\Omega'([v],[u]) = 0$ for all $u \in G$ then $\Omega(v,u) = 0$ for all $u \in G$ then $v \in G^\perp \implies [v] = 0$. So $(W',\Omega')$ is a symplectic vector space.
Since $L \cap G^\perp = \{0\}$ we have that $L' = L\cap G \rightarrow G/G^\perp = W'$ is injective, so $L'$ embedded as a subspace in $W'$. For any $[v],[u] \in L'$, we can represents them uniquely with $v, u \in L$, so $\Omega'([v],[u]) = \Omega(v,u) = 0$. To show that $L'$ is Lagrangian we compute the dimension \begin{equation} \dim L' = \dim G + \dim L - \dim W = \dim G + \frac{1}{2}\dim W - \dim W = \dim G - \frac{1}{2}\dim W. \end{equation} On the other hand, we have \begin{equation} \dim W' = \dim G - \dim G^\perp. \end{equation} We also know that $\dim G + \dim G^\perp = \dim W$ so \begin{equation} \dim W'= 2\dim G - \dim W. \end{equation} Therefore, $\dim L' = \frac{1}{2}\dim W'$ which implies that $L'$ is Lagrangian.
As you can see my argument used a lot of dimension counting which won't work when $W$ is infinite dimensional. So I'm wondering if there is a more general argument which would extend the result to the case where $W$ is an infinite dimensional symplectic vector space? Possibly with an extra assumption that $W' = G/G^\perp$ is finite dimensional?