As William Elliot noted, the initial statement was false. I changed it to a weaker one.
Let $L$ be a distributive lattice, denote $lu(x)$ - a least upper bound of elements which cover $x$ (to be clear: $y$ covers $x$ if $x < y$ and there is no element $z$ such that $x < z < y$), $gl(x)$ - a greatest lower bound of set of elements covered by $x$.
Prove (or disprove): let $x, y \in L$ such that $x \lor y = lu(x)$ (least upper bound of $x$ and $y$ equals to $lu(x)$). Then $gl(y) \le x$. Would be grateful for any hint.
Presumably the lattice is a complete lattice because the covers of an element may be an infinite set. There is a counter example.
Consider the complete distributive lattice [0,1].
There are no covers of 1.
So the least upper bound of the covers is 0.
As 0 does not cover any element
the greatest lower bound of the elements 0 covers is 1.