In Munkres's Topology, it is asked to prove that, if a topological space $X$ is compact Hausdorff, then to every open covering $\mathcal A$ of $X$, there exists a finer open covering $\mathcal B$ with the following property:
If $B_1,B_2 \in \mathcal B$ have non-empty intersection, then $B_1 \cup B_2$ lies in an element of $\mathcal A$.
There is a hint in the question that suggests to take a finite subcovering $A_1,\ldots,A_n$, choose an open covering $C_1,\ldots,C_n$ of $X$ such that $\overline C_i \subset A_i$ for each $i$, then consider the sets $B_J = \bigcap\limits_{j \in J} A_j - \bigcup\limits_{j \notin J} \overline C_j$ with $\varnothing \neq J \subset \{1,\ldots,n\}$.
It is not clear to me that such $B_J$ satisfy the required property. Furthermore, I don't see how this can lead to an open covering that refines $\mathcal A$. Could someone shed a light on this?