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 $G\_{\cal C} \cong G$?



Representing graphs of diameter $2$ with full intersection graphs
The answer is no.
Let $G$ be the four-cycle $abcd$. Suppose there was a ground-set $X$ and a *full* family of sets $A, B, C, D$ such that $A$ represents $a$, $B$ represents $b$, etc.
Since $ab \in E(G)$, there is some $x \in A \cap B$. As $cd \in E(G)$, there is some $y \in C \cap D$. 


* $ac \notin E(G)$, so $x \notin C$ and $y \notin A$.
* $bd \notin E(G)$, so $x \notin D$ and $y \notin B$.


Therefore, none of the four sets contains both $x$ and $y$. The family is not full.


