A metric space with the Haver property whose square fails this property

Haver introduced the following property of metric spaces $(X,d)$: for each sequence $\epsilon _{1}, \epsilon _{2}, \ldots$ of positive numbers there exist collections $\mathcal{V}_{1}, \mathcal{V}_{2}, \ldots$ of open subsets of $X$, the union $\bigcup _{i}\mathcal{V}_{i}$ of which covers $X$, such that the members of $\mathcal{V}_{i}$ are pairwise disjoint and every member of $\mathcal{V}_{i}$ has diameter less than $\epsilon _{i}$. We construct two separable complete metric spaces $(X_{0},d_{0})$, $(X_{1},d_{1})$ with the Haver property such that $d_{0}$, $d_{1}$ generate the same topology on $X_{0}\cap X_{1}\neq \emptyset$, but $(X_{0}\cap X_{1}, \max (d_{0},d_{1}))$ fails this property. In particular, the square of a separable complete metric space with the Haver property may fail this property. Our results answer some questions posed by Babinkostova in 2007.