= Solution
For a <connected intersection of a closed filter base>, the filter-base property implies the <finite intersection property>. <Compactness>, in its closed-set formulation, therefore gives $K=\bigcap\mathcal F\ne\varnothing$. It is closed in $X$, hence <compact>.
Suppose additionally that each base member is <connected> but $K$ is disconnected. Write $K=A\cup B$ with $A,B$ disjoint nonempty <closed subsets> of $K$. They are <compact>, hence closed in the <Hausdorff space> $X$. <Normality of a compact Hausdorff space> gives disjoint <open sets> $U,V$ containing $A,B$, respectively. The <compact set> $W=X\setminus(U\cup V)$ misses $K$. For every $w\in W$ some $F_w\in\mathcal F$ misses $w$, so the <open sets> $X\setminus F_w$ cover $W$. A finite subcover gives $F_1\cap\cdots\cap F_m\subseteq U\cup V$. Choose $F_0\in\mathcal F$ inside this <finite intersection>. If $W$ is empty, choose any base member instead.
Since $F_0$ contains $K$, it meets both $U$ and $V$; but it lies in their <disjoint union>. These two relatively open nonempty pieces disconnect $F_0$, a contradiction. Thus \b[the <intersection> is nonempty and <connected>]. <Compactness> is used to pass from an infinite exclusion condition to a single base member.
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