Here is a proof that is based on Brook's Theorem. We plagiarize you and start by noting that $r(C_4,K_3)=7$, and so $G$ has minimum degree at least 3. We then plagiarize Chvatal, Harary and note that $G$ has chromatic number at least 4. Thus, by Brook's Theorem, $G$ has maximum degree $\Delta(G)$ at least 4. Let $v$ be a vertex of maximum degree and let $N(v)$ be the neighbours of $v$ and let $S(v)$ be the non-neighbours of $v$. Since $G$ has no $C_4$, note that each vertex in $S(v)$ has at most one neighbour in $N(v)$. Thus, the minimum degree of the subgraph induced by $S(v)$ is at least 2. This rules out $\Delta(G)=9,8$, or $7$.
If $\Delta(G)=6$, then there are at least three vertices $x,y,z \in N(v)$ which are not adjacent to any vertex in $S(v)$. Since $x$ has degree at least 3 in $G$, it must be adjacent to at least two other vertices in $N(v)$, which creates a $C_4$.
If $\Delta(G)=5$, then $G[S(v)]$ is a graph on 4 vertices with minimum degree 2. Such a graph necessarily contains a $C_4$.
We now suppose $\Delta(G)=4$. In this case, $G[S(v)]$ is a graph on 5 vertices with minimum degree 2. Thus, every cycle of $G[S(v)]$ must be of length 3 or 5. If $G[S(v)]$ contains a $C_5$, then $G[S(v)]=C_5$, since adding any chord to a $C_5$ produces a $C_4$. This implies that some Thus, each vertex in $G[S(v)]$ has exactly one neighbour in $N(v)$. Hence, $G[N(v)]$ must be a matching ${ab, cd}$ of size 2, else $C$ G$has a vertex of degree 2. It follows that each vertex in$G$, since N(v)$, has at least one neighbour in $N(v)$ only contains four S(v)$. Thus, one vertex (say$a$) has exactly two neighbours in$S(v)$, while$b,c$, and$d$have exactly one neighbour in$S(v)$. Let$x,y,z$be the vertices in$S(v)$which are not adjacent to either$b$or$c$. If any of$xy,yz,xz \notin E(G)$, then$G$has a contradictionstable set of size 4. It follows Thus,$xyz$is a triangle. This contradicts that$G[S(v)]=C_5$. The only remaining possibility is that$G[S(v)]$must be is a bowtie. In particular,$G$has two non-adjacent vertices$u$and$v$of degree 4 such that$N(u)$and$N(v)$are disjoint. Thus, the subgraph$H$of$G$induced by$N(u) \cup N(v)$has minimum degree at least 2. In particular$H$contains a cycle$C$. It is easy to verify that if$|C|=3,4,5,6,$or$7$, then$G$contains a$C_4$since$H$cannot contain a vertex with two neighbours in$N(u)$or two neighbours in$N(v)$. Thus,$|C|=8$. But then$H=C_8$else$H$contains a cycle of smaller length. Thus, every other vertex of$H$is an independent a stable set in$G$, a contradiction. Note that this proof avoids the use of the minimum degree lemma. 2 added 17 characters in body It seems as if the Chvatal, Harary proof has a logical gap, and your proof seems to be missing some details. Here is a proof that is based on Brook's Theorem. We plagiarize you and start by noting that$r(C_4,K_3)=7$, and so$G$has minimum degree at least 3. We then plagiarize Chvatal, Harary and note that$G$has chromatic number at least 4. Thus, by Brook's Theorem,$G$has maximum degree$\Delta(G)$at least 4. Let$v$be a vertex of maximum degree and let$N(v)$be the neighbours of$v$and let$S(v)$be the non-neighbours of$v$. Since$G$has no$C_4$, note that each vertex in$S(v)$has at most one neighbour in$N(v)$. Thus, the minimum degree of the subgraph induced by$S(v)$is at least 2. This rules out$\Delta(G)=9,8$, or$7$. If$\Delta(G)=6$, then there are at least three vertices$x,y,z \in N(v)$which are not adjacent to any vertex in$S(v)$. Since$x$has degree at least 3 in$G$, it must be adjacent to at least two other vertices in$N(v)$, which creates a$C_4$. If$\Delta(G)=5$, then$G[S(v)]$is a graph on 4 vertices with minimum degree 2. Such a graph necessarily contains a$C_4$. The only remaining possibility is$\Delta(G)=4$. In this case,$G[S(v)]$is a graph on 5 vertices with minimum degree 2. Thus, every cycle of$G[S(v)]$must be of length 3 or 5. If$G[S(v)]$contains a$C_5$, then$G[S(v)]=C_5$, since adding any chord to a$C_5$produces a$C_4$. This implies that$G$has a some vertex of$C$has degree 2 in$G$, since$N(v)$only contains four vertices, a contradiction. Therefore It follows that$G[S(v)]$must be a bowtie. In particular,$G$has two non-adjacent vertices$u$and$v$of degree 4 such that$N(u)$and$N(v)$are disjoint. Thus, the subgraph$H$of$G$induced by$N(u) \cup N(v)$has minimum degree at least 2. In particular$H$contains a cycle$C$. It is easy to verify that if$|C|=3,4,5,6,$or$7$, then$G$contains a$C_4$since$H$cannot contain a vertex with two neighbours in$N(u)$or two neighbours in$N(v)$. Thus,$|C|=8$. But then$H=C_8$else$H$contains a cycle of smaller length. Thus, every other vertex of$H$is an independent set in$G$, a contradiction. 1 It seems as if the Chvatal, Harary proof has a logical gap, and your proof seems to be missing some details. Here is a proof that is based on Brook's Theorem. We plagiarize you and start by noting that$r(C_4,K_3)=7$, and so$G$has minimum degree at least 3. We then plagiarize Chvatal, Harary and note that$G$has chromatic number at least 4. Thus, by Brook's Theorem,$G$has maximum degree$\Delta(G)$at least 4. Let$v$be a vertex of maximum degree and let$N(v)$be the neighbours of$v$and let$S(v)$be the non-neighbours of$v$. Since$G$has no$C_4$, note that each vertex in$S(v)$has at most one neighbour in$N(v)$. Thus, the minimum degree of the subgraph induced by$S(v)$is at least 2. This rules out$\Delta(G)=9,8$, or$7$. If$\Delta(G)=6$, then there are at least three vertices$x,y,z \in N(v)$which are not adjacent to any vertex in$S(v)$. Since$x$has degree at least 3 in$G$, it must be adjacent to at least two other vertices in$N(v)$, which creates a$C_4$. If$\Delta(G)=5$, then$G[S(v)]$is a graph on 4 vertices with minimum degree 2. Such a graph necessarily contains a$C_4$. The only remaining possibility is$\Delta(G)=4$. In this case,$G[S(v)]$is a graph on 5 vertices with minimum degree 2. Thus, every cycle of$G[S(v)]$must be of length 3 or 5. If$G[S(v)]$contains a$C_5$, then$G[S(v)]=C_5$, since adding any chord to a$C_5$produces a$C_4$. This implies that$G$has a vertex of degree 2 since$N(v)$only contains four vertices, a contradiction. Therefore$G[S(v)]$must be a bowtie. In particular,$G$has two non-adjacent vertices$u$and$v$of degree 4 such that$N(u)$and$N(v)$are disjoint. Thus, the subgraph$H$of$G$induced by$N(u) \cup N(v)$has minimum degree at least 2. In particular$H$contains a cycle$C$. It is easy to verify that if$|C|=3,4,5,6,$or$7$, then$G$contains a$C_4$since$H$cannot contain a vertex with two neighbours in$N(u)$or two neighbours in$N(v)$. Thus,$|C|=8$. But then$H=C_8$else$H$contains a cycle of smaller length. Thus, every other vertex of$H$is an independent set in$G\$, a contradiction.