(This is a follow-up to [this](https://mathoverflow.net/questions/265422/walking-withouth-gaps-through-a-set-of-sets) question.)


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. all members of ${\cal C}$ contain at least $2$ elements, and $X\notin {\cal C}$;
2. $A\neq B\in {\cal C}$ implies $|A\cap B| \leq 1$; and
3. 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 $f(k)\cap f(k+1)\neq \emptyset$?



A walk through a set of sets
Yes. For any edge $(x,y)$ of a complete graph on $X$ there exists unique set $A(x,y)$ in $\cal C$ containing both $x,y$. Thus if we find a Hamiltonian path in the line graph of $K\_m$ (that is pretty easy to do), it produces a Hamiltonian path for $\cal C$. 


