Covering property 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?
 A: Following the hint, let $J$ be a non-empty subset of $\{1,\dots,n\}$. Then $\bigcap\limits_{j\in J}A_j$ is open, and $\bigcup\limits_{j\notin J}\overline C_j$ is closed, so $B_J=\bigcap\limits_{j\in J}A_j\setminus\bigcup\limits_{j\notin J}\overline C_j$ is open. Since $J\ne\varnothing$, there is some $j_0\in J$, and clearly $B_J\subseteq\bigcap\limits_{j\in J}A_j\subseteq A_{j_0}$, so $\mathcal{B}=\big\{B_J:\varnothing\ne J\subseteq\{1,\dots,n\}\big\}$ refines $\mathcal{A}$.
To show that $\mathcal{B}$ covers $X$, let $x\in X$, and let $J=\big\{k\in\{1,\dots,n\}:x\in A_k\}$; clearly $x\in\bigcap\limits_{j\in J}A_j$ and $x\notin \bigcup\limits_{j\notin J}A_j\supseteq\bigcup\limits_{j\notin J}\overline C_j$, so $x\in B_J$.
It only remains to show that if $B_J,B_K\in\mathcal{B}$ and $B_J\cap B_K\ne\varnothing$, then there is some $A\in\mathcal{A}$ such that $B_J\cup B_K\subseteq A$. Fix $x\in B_J\cap B_K$; there is some $j\in\{1,\dots,n\}$ such that $x\in C_j$. Suppose that $j\notin J$; then $B_J\cap \overline C_j=\varnothing$ (by the definition of $B_J$), so $x\notin B_J$, contradicting the choice of $x$. Therefore $j\in J$, and hence $B_J\subseteq A_j$. Exactly the same argument shows that $j\in K$ and therefore $B_K\subseteq A_j$, and we’re done.
