Let $(L,\land,\lor)$ be a complete distributive lattice. Given $x\neq y \in L$, is there a finite set ${\cal I}$ of closed intervals in $L$ such that


1. no member of ${\cal I}$ contains both $x$ and $y$, and
2. $\bigcup {\cal I} = L$


?


(A *closed interval* in $L$ is a subset of the form $[a, b] = \{x\in L: a\leq x\leq b\}$ where $a\leq b \in L$.)



Covering property of complete distributive lattices
I think you do mean *completely* distributive, not just finitely. Otherwise $\mathbb{Z}$ with its usual ordering is not a finite union of *any* set of closed intervals. For complete distributive lattices, let $L$ be the lattice of all measurable subsets of the unit interval $[0,1]$ modulo sets of measure 0, ordered by inclusion (up to sets of measure 0). This a complete distributive lattice (in fact, a complete boolean algebra). It is easy to see that if $L$ is written as a finite union of closed intervals, then one of these intervals is the whole lattice.


