Open problem: co-separability of filter objects

Conjecture Let $latex a$ and $latex b$ are filters on a set $latex U$. Then

$latex a\cap b = \{U\} \Rightarrow \\ \exists A,B\in\mathcal{P}U: (\forall X\in a: A\subseteq X \wedge \forall Y\in b: B\subseteq Y \wedge A \cup B = U).$ [corrected]

This conjecture can be equivalently reformulated in terms of filter objects:

Conjecture Let $latex \mathcal{A}$ and $latex \mathcal{B}$ are filter objects on a set $latex U$. Then

$latex \mathcal{A}\cup^{\mathfrak{F}}\mathcal{B}=U \Rightarrow \exists A,B\in\mathcal{P}U: (A\subseteq\mathcal{A} \wedge B\subseteq\mathcal{B} \wedge A\cup B=U).$

(where $latex \mathfrak{F}$ is the set of filter objects on $latex U$).

It is simple to show (by applying the above conjecture twice) that it is equivalent to a yet simpler conjecture

Conjecture Let $latex \mathcal{A}$ and $latex \mathcal{B}$ are filter objects on a set $latex U$. Then

$latex \mathcal{A}\cup^{\mathfrak{F}}\mathcal{B}=U \Rightarrow \exists A\in\mathcal{P}U: (A\subseteq\mathcal{A} \wedge A\cup^{\mathfrak{F}}\mathcal{B}=U).$

Put in yet simpler words, this conjecture can be formulated: the filtrator of filters on a set is with co-separable center.

Looking innocent? Indeed I currently don’t know how to attack this looking simple problem. Maybe you will help?