Let $X\neq \emptyset $ be a finite set and suppose that ${\cal C}$ is a set of subsets of $X$ with the following properties:


1. $X\notin {\cal C}$, and
2. for all $x,y\in X$ there is $A\in {\cal C}$ such that $\{x,y\}\subseteq A$.


Let $m=|{\cal C}|$. Is there a bijection $f: \{1,\ldots, m\}\to {\cal C}$ such that for all $k \in \{1,\ldots, m-1\}$ we have $f(k)\cap f(k+1)\neq \emptyset$?



Walking "withouth gaps" through a set of sets
Still no. You may choose three subsets which cover all pairs of elements and add many mutually disjoint subsets to this collection. 


