A split set with both a complete and an anticomplete outside vertex yields a nontrivial module

First claim: if $x\in V(G)\setminus S$ is neither complete nor anticomplete to $S$, then either $c$ is adjacent to $a$ and $x$ is adjacent to $c$, or $c$ is nonadjacent to $a$ and $x$ is nonadjacent to $a$. Indeed, let $S_1$ be the neighbors of $x$ in $S$ and $S_2=S\setminus S_1$; both are nonempty. By [F1], either there are $u,v\in S_1$ and $w\in S_2$ with $u$-$v$-$w$ an induced path, or there are $u\in S_1$ and $v,w\in S_2$ with $uw\in E(G)$ while $uv,vw\mathrm{\notin }E(G)$. In the first case bull-freeness rules out both $x$-$a$ and $\overline{xc}$ simultaneously, because otherwise $\{a,x,u,v,w\}$ and then $\{x,v,w,c,a\}$ would be bulls. In the second case bull-freeness similarly rules out both $\overline{xc}$ and $x$-$a$, because otherwise $\{x,u,w,c,v\}$ and then $\{a,x,u,c,v\}$ would be bulls.

Let $C$ be the set of vertices complete to $S$, let $A$ be the set of vertices anticomplete to $S$, and let $X=V(G)\setminus (S\cup C\cup A)$. Either every vertex of $C$ has a neighbor in $A$, or every vertex of $A$ has a nonneighbor in $C$: otherwise a vertex of $C$ anticomplete to $A$ and a vertex of $A$ complete to $C$ would contradict each other. Replacing $G$ by $\bar{G}$ if necessary preserves bull-freeness, splitness, and modules by [L1], [F1], and [F2], so assume that every vertex of $C$ has a neighbor in $A$. Step 1.1 then makes $C$ complete to $X$. Let $A^{\mathrm{\prime }}$ be the set of vertices of $A$ lying on an induced path $x$-$a_1$-$\cdots$-$a_k$ with $x\in X$ and all $a_i\in A$. We prove by induction on $k$ that $a_k$ is complete to $C$. For $k=1$, if $c^{\mathrm{\prime }}{a}_1$ were a nonedge for some $c^{\mathrm{\prime }}\in C$, step 1.1 applied to $x,c^{\mathrm{\prime }},a_1$ would force $x{a}_1$ to be a nonedge, a contradiction. For $k>1$, put $a_0=x$ and assume the result through $a_{k-1}$. Choose $s\in S$ nonadjacent to $a_{k-2}$ when $k=2$, which is possible because $x$ is mixed on $S$; for $k>2$ any $s\in S$ works because $a_{k-2}\in A$. If some $c^{\mathrm{\prime }}\in C$ were nonadjacent to $a_k$, then $\{s,c^{\mathrm{\prime }},a_{k-2},a_{k-1},a_k\}$ would be a bull: $c^{\mathrm{\prime }},a_{k-2},a_{k-1}$ form its triangle, while $s$ and $a_k$ are pendant at $c^{\mathrm{\prime }}$ and $a_{k-1}$. Hence $A^{\mathrm{\prime }}$ is complete to $C$.

Put $Z=S\cup X\cup A^{\mathrm{\prime }}$. Every vertex of $A\setminus A^{\mathrm{\prime }}$ is anticomplete to $Z$: it is anticomplete to $S$ by definition, and any path from it to $X\cup A^{\mathrm{\prime }}$ through $A$ has a shortest, hence induced, subpath that would put it in $A^{\mathrm{\prime }}$. Every vertex of $C$ is complete to $Z$ by step 2.1 and the definition of $C$. Thus every outside vertex is complete or anticomplete to $Z$, so $Z$ is a module by [F2]. The set $Z$ is nontrivial because $S\subseteq Z$ and $\mathrm{\mid }S\mathrm{\mid }>1$, while $c\in C\setminus Z$, so $Z\mathrm{\ne }V(G)$.