Verbality is not finite-intersection-closed

Statement
It is possible to have a group $$G$$ and verbal subgroups $$H,K$$ of $$G$$ such that the intersection of subgroups $$H \cap K$$ is not a verbal subgroup of $$G$$.

Proof
Consider $$F$$ to be the free nilpotent group on three generators $$a,b,c$$.


 * Let $$G$$ be the central product of $$F$$ with $$\langle d \rangle$$ (a copy of $$\mathbb{Z}$$ where $$d^2$$ is identified with $$[a,b]$$.
 * Let $$H$$ be the derived subgroup $$G'$$, so that $$H$$ is a verbal subgroup of $$G$$.
 * Let $$K$$ be the subgroup of $$G$$ generated by squares, so that $$K$$ is a verbal subgroup of $$G$$.
 * The subgroup $$H \cap K$$ is the subgroup generated by the element $$d^2$$ and the squares of all elements in $$G'$$. It can be verified that this is not a verbal subgroup of $$G$$.