We prove that if a family of compact connected sets in the plane has the property that every three members of it are intersected by a line, then there are three lines intersecting all the sets in the family. This answers a question of Eckhoff from 1993, who proved that, under the same condition, there are four lines intersecting all the sets. In fact, we prove a colorful version of this result, under weakened conditions on the sets. A triple of sets $A,B,C$ in the plane is said to be a {em tight} if $textrm{conv}(Acup B)cap textrm{conv}(Acup C)cap textrm{conv}(Bcap C) eq emptyset.$ This notion was first introduced by Holmsen, where he showed that if $mathcal{F}$ is a family of compact convex sets in the plane in which every three sets form a tight triple, then there is a line intersecting at least $frac{1}{8}|mathcal{F}|$ members of $mathcal{F}$. Here we prove that if $mathcal{F}_1,dots,mathcal{F}_6$ are families of compact connected sets in the plane such that every three sets, chosen from three distinct families $mathcal{F}_i$, form a tight triple, then there exists $1le jle 6$ and three lines intersecting every member of $mathcal{F}_j$. In particular, this improves $frac{1}{8}$ to $frac{1}{3}$ in Holmsens result.