Remove two vertices such that there are no 3-cliques in the resulting graph

Question

The question is: A graph G contains no 5-cliques and every two 3-cliques intersect in at least one vertex. Show that we can delete two vertices such that the resulting graph contains no 3-cliques.

I tried looking at ways of drawing 3-cliques such that every two intersect but more of them do not intersect in the same two points. The 5-clique is obviously a counter-example, but how can I show that if there is no five clique and all 3-cliques are non-disjoint then all of them contain (at least) one of two points {a,b}?

Maybe by looking at the 3-cliques as sets of vertices and then looking at the intersections between pairs of 3-cliques? But it only needs to contain one of the two points, so the intersection is too strong.

Answer 1

Assume, on the contrary, that there exists a graph $G$, $|V(G)|>1$ in which
(i) any two triangles have at least one vertex in common;
(ii) there are no $5$-cliques;
(iii) for any two vertices of $V(G)$, there exists a triangle without these two vertices.

In view (iii) and $|G|>1$, there exists triangles $abc$ and $axy$ in $G$ and the vertices $a,b,c,x,y$ are pairwise distinct (why?).

In view (ii) $abcxy$ is not a clique. Without loss of generality we can assume that $bx\notin E(G)$. Hence again in view of (iii), (i) for the pair of vertices $a,c$ there exists a triangle $buz$ in $G$. If $\{u,z\}\cap\{x,y\}=\varnothing$, then triangles $axy$ and $buz$ have no common vertices, which is impossible by (i). Let $u=y$. Then $z\notin\{x,y\}$.
Thus we have the subgraph of $G$ shown in Figure 1.

By (i) and (iii) for a pair $a,y$ there exists a triangle $bxw$ for some $w\in V(G)$.
But this is impossible, since according to our assumption $bx\notin E(G)$.