Let $G_1, G_2$ and $G_3$ be undirected simple graphs such that $(G_1\cap G_2)\cup(G_2\cap G_3)\cup(G_1\cap G_3)=G_2$. How to show that $G_1\subseteq G_2\subseteq G_3$?
Note that the graph's union $\cup$ and intersection $\cap$ act as lattice operations $\vee$ and $\wedge$, respectively. Also, $\cap$ distributes over $\cup$.
Edited Also, $G_1, G_2$ and $G_3$ have the same vertex set.
An easy way to see that this cannot be true is to observe that your hypothesis does not change if you swap $G_1$ and $G_3$. Thus if your conjecture were true, it would not only imply $G_1 \subseteq G_2 \subseteq G_3$, but also $G_3 \subseteq G_2 \subseteq G_1$ and hence $G_1 = G_3$.