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 $\textrm{conv}(A\cup B)\cap \textrm{conv}(A\cup C)\cap \textrm{conv}(B\cap C)\neq \emptyset.$ 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 $1\le j\le 6$ and three lines intersecting every member of $\mathcal{F}_j$. In particular, this improves $\frac{1}{8}$ to $\frac{1}{3}$ in Holmsen's result.