Let $H$ be a separable Hilbert space. Let $C(H)$ be the set of closed
subspaces of the Hilbert space $H$.
Let $\le$ be the inclusion $\subseteq$. For $a,b \in C(H)$ define
$a^\perp$ as the othogonal closed space of $a$,
$a\wedge b = a \cap b $,
$a \vee b= \{$ the smallest closed subspace that contains $a$ and $b \}$.
Using these definitions, $C(H)$ is an orthomodular lattice, and it will be modular if and only if $H$ has finite dimension.
N.B. A lattice $\mathcal L$ is said to be modular iff $$\forall a,b \in \mathcal L: a\le b \quad\Rightarrow \quad\forall c \in \mathcal{L}: a \vee (c \wedge b) = (a \vee c) \wedge b.$$
I'd like help with finding an example of three closed subspaces $a, b, c$ of an infinite dimensional Hilbert space $H$ such that $a\le b$ doesn't imply $a \vee (c \wedge b) = (a \vee c) \wedge b $.