If ${\cal C}$ is a collection of subsets of a set $X$, we associate to ${\cal C}$ a graph $G_{\cal C} = (V,E)$ where $V = {\cal C}$ and $$E = \big\{\{A,B\}: A\neq B\in {\cal C} \land A\cap B \neq \emptyset\big\}.$$

We call ${\cal C}$ *full on $X$* if for all $x,y\in X$ there is $A\in {\cal C}$ such that $\{x,y\} \subseteq A$.

Let $G$ be a simple, undirected graph such that for any $2$ distinct vertices, there is a path of length at most $2$ connecting them. Is there a set $X$ and a full collection of subsets ${\cal C}$ on $X$ such that $G_{\cal C} \cong G$?