Let $X$ be a set and $\mathcal{F}$ be a finite family of subsets of $X$. We call a set $T\subseteq X$ a tranversal set for $\mathcal{F}$ if it intersects every set in $\mathcal{F}$.
Suppose that $X=\mathbb{R}$ and $\mathcal{F}$ is a finite family of intervals such that any two intersect. Then it is easy to prove that $\mathcal{F}$ admits a transversal set with only one element, namely $\cap\mathcal{F}\neq \emptyset$.
Suppose that every set in $\mathcal{F}$ is union of at most $n$ points and intervals in $\mathbb{R}$, and again that every two sets in $\mathcal{F}$ intersect. Is there a simple combinatorial argument to show whether $\mathcal{F}$ admits a finite transversal set of a certain cardinality?
My guess here being that one such family might always admit a transversal set of cardinality $n$.