 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 the intersection is nonempty and connected. Compactness is used to pass from an infinite exclusion condition to a single base member.