Let $\mathfrak{A}$ be a poset, let $a\in\mathfrak{A}$. By definition $$\star a = \{ x\in\mathfrak{A} \mid \text{there exists non-least } y\in\mathfrak{A} \text{ such that } y\le a \text{ and } y\le x\}.$$
I call a poset $\mathfrak{A}$ strongly separable, when $\star a\subseteq\star b \Rightarrow a\le b$ for every $a,b\in\mathfrak{A}$.
Now fix a family $\mathfrak{A}_{i\in n}$ of posets. Under which conditions the product order $\prod\mathfrak{A}$ for this family of posets is strongly separable? (We assume that every poset $\mathfrak{A}_i$ is strongly separable.)
Sorry, I would like to make a comment but unfortunately I don't have enough reputation.
If $\mathfrak{A}_i$ is strongly separable and it is not lower bounded then $\prod\mathfrak{A}_i$ is strongly separable.
To simplify let's consider the case with n=2.
Suppose that there is $(a,a'),(b,b') \in \mathfrak{A}_1 \times \mathfrak{A}_2$ such that $\star (a,a') \subseteq \star (b,b')$. We claim that $\star a \subseteq \star b$ and $\star a' \subseteq \star b'$.
Let $x \in \star a$. Then $(x,a') \in \star (a,a')$. Since $ \star (a,a') \subseteq \star (b,b')$ we have that $(x,a') \in \star (b,b')$. In particular this means that there exists $z \in \mathfrak{A}_1$ such that $z \leq x$ and $z \leq b$. Given that $\mathfrak{A}_1$ is not lower bounded, we conclude that $x \in \star b$.