Let $G=(V,E)$ be a Split Graph with $|V| \geq 4$. Then how to prove that:
No induced sub-graph of G with 4 Vertices is a cycle with length 4 OR a pair of not incident edges?
Well it must be from Földes, Stéphane; Hammer, Peter L. (1977a), "Split graphs". So
A graph is split if and only if no induced subgraph is a cycle on four or five vertices, or a pair of disjoint edges (the complement of a 4-cycle).
But how to prove it?
Let $A$ be the "clique" part of the split graph and $B$ the "coclique" part, i.e., the subgraphs induced by $A$ and $B$ are a complete graph and independent set respectively. Write $C_4$ and $K_2+ K_2$ for the two proposed forbidden subgraphs.
Now choose four vertices $a,b,c,d$ and let $H$ be the subgraph induced by them. If three of $a,b,c,d$ are in $A$, then $H$ contains a triangle and so $H\neq C_4, K_2+K_2$. If three of $a,b,c,d$ are in $B$, say $a,b,c$, then any edges of $H$ must be incident to $d$. So again $H\neq C_4,K_2+K_2$ since both $C_4$ and $K_2+K_2$ have disjoint edges.
The final possibility is that exactly two of $a,b,c,d$ are in $A$, say $a,b$. So $a$ and $b$ are adjacent. But then if $H=C_4$ or $K_2+K_2$, we must have $c$ and $d$ adjacent which is impossible since they are in $B$.
So no induced subgraph on four vertices of $H$ can be equal to $C_4$ or $K_2+K_2$.