Left side contains two cycles $aceg$ and $hbdf$ of length $4$, with added property that within these cycles, non-adjacent vertices are connected by an edge ($gc$ and $ae$ for $aceg$). But there is no such cycles in the right image. But I'm afraid this isn't going to be enough of a proof.
Any guidance?
Your argument looks right to me, and it isn't hard to extend it to a convincing proof. As you've pointed out, every vertex in the left graph belongs to one of two isomorphic copies of $K_4,$ and there's no $K_4$ on the right side. For a proof, note that if the right-hand graph is isomorphic to the left-hand one, then $p$ belongs to some $K_4$. The neighbors of $p$ are $q,s,v, \text{ and } w.$ Which of these can be in a $K_4$ along with $p$? Not $s,$ which is adjacent to $r$ and $t$. Not $q,$ which is adjacent to $r$ and $u$. Now we're done, because there aren't enough vertices left to make a $K_4$ containing $p$.