Let $(V,\Omega)$ be a symplectic vector space. A subspace $L$ is called Lagrangian if $L^\perp=L$. A subspace $W$ is called isotropic if $W\subset W^\perp$.
For any isotropic subspace $W$ there is a Lagrangian subspace $W$ such that $W\subset L$
This is easy to see if we let $e_1,...,e_k$ be a basis of $W$ and extend this to a symplectic basis of $V$. Then we can take $L=span\{e_1,...,e_n\}$. I would like to prove this without resorting to a basis to get used to working with the machinery.
I think it is important here to assume that the pertinent vector spaces are finite dimensional; otherwise the statement is false. I proceed to a proof.
Let $W$ be isotropic. If there exists an isotropic subspace $W'$ containing $W$ properly ($W \subsetneq W'$), we replace $W$ by $W'$. We repeat inductively after there exists no such $W'$, i.e. until we obtain a maximal isotropic subspace. This always happens after finitely many steps, because the dimensions of consecutive subspaces grows and is bounded by $\dim(V) < \infty$. This proves that every isotropic subspace is contained in a maximal isotropic subspace.
The proof will be completed by showing that every maximal isotropic subspace is Lagrangian (the converse is also true; I suggest showing that as an exercise). Suppose that $W$ is maximal isotropic. We have to show that $W^{\perp} \subset W$. For the sake of contradiction assume that $x$ is an element of $W^{\perp}$ which is not in $W$. Then the linear span of $x$ and $W$ is an isotropic subspace properly containing $W$, which is absurd because $W$ is maximal.