Let $G=(V,E)$ be a finite simple graph. Does there exist a finite set $S$ and $S_G\subseteq P(S)$ such that the intersection graph $H$ of $S_G$ is isomorphic to $G$, i.e. $H\cong G$.
Here's my idea:
If I can find such an intersection graph $H$ that $H\cong G$, then the proposition is proved.
Let $V=\{v_1,\dots,v_n\}$ and $E=\{e_1,\dots,e_m\}$, where $n,m$ are both positive integers since $G$ is finite. Now, let $S=V$, and $S_G=E\cup \{\{v_i\}\}$, where $v_i$ satisfies the condition that $v_i\neq e_j\cap e_k$ for all $j,k\le m$. That is, $v_i$ is a vertex with degree 1. We define the set of edges $F$ by
$$f=\begin{cases}\{v_i,e_i\}&\text{ if }v_i\in S_G\ \text{and}\ v_i\text{ is incident with }e_i\\\{e_j,e_k\}&\text{ for all }e_j,e_k\in E\text{ with }j\neq k\end{cases}$$
Then $H=(S_G,F)\cong G$.
I find my idea works well for closed graph, but not so well for others. There must be some problems in my idea. Anyone has any idea or can modify my solution?