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 some vertex of $C$ has degree 2 in $G$, since $N(v)$ only contains four vertices, a contradiction. 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.

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 $C_4$.

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 $C_4$. 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 $G$ has a vertex of degree 2. It follows that each vertex in $N(v)$, has at least one neighbour in $S(v)$. 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 stable set of size 4. Thus, $xyz$ is a triangle. This contradicts that $G[S(v)]=C_5$.

The only remaining possibility is that $G[S(v)]$ 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 a stable set in $G$, a contradiction.

Note that this proof avoids the use of the minimum degree lemma.

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.

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 some vertex of $C$ has degree 2 in $G$, since $N(v)$ only contains four vertices, a contradiction. 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.