For a vertex in a basic bull-free graph, either its neighborhood or its antineighborhood is perfect

Suppose neither $G[N]$ nor $G[M]$ is perfect. First they cannot both contain odd holes. Indeed, let $H_N\subseteq G[N]$ and $H_M\subseteq G[M]$ be odd holes of lengths $n$ and $m$. Every vertex of $H_M$ is nonadjacent to $u$, while $u$ is complete to $H_N$, so [L1] gives each vertex of $H_M$ at least $n-2$ neighbors in $H_N$. Thus there are at least $m(n-2)$ cross edges. On the other hand every vertex of $H_N$ is adjacent to $u$, while $u$ is anticomplete to $H_M$, so [L2] gives each vertex of $H_N$ at least $(m+1)\mathrm{/}2$ nonneighbors in $H_M$. Hence there are at least $n(m+1)\mathrm{/}2$ cross nonedges. Since there are only $mn$ cross pairs altogether, we obtain $m(n-2)+n(m+1)\mathrm{/}2\le mn$, equivalently $mn-4m+n\le 0$, impossible because $m,n\ge 5$ make the left-hand side at least $m+5>0$.

By [L4], the same argument in $\bar{G}$ shows that $G[N]$ and $G[M]$ cannot both contain odd antiholes. If $G[N]$ contained an odd hole, then step 1.1 would force $G[M]$ to contain no odd hole, so [L3] would give an odd antihole $H_M$ in $G[M]$. Because a $5$-antihole is also a $5$-hole, step 1.1 excludes the case $\mathrm{\mid }{H}_M\mathrm{\mid }=5$, and the same complement argument excludes $\mathrm{\mid }{H}_N\mathrm{\mid }=5$; hence both have length at least $7$. Now every vertex of $H_M$ is nonadjacent to $u$, so [L1] applied to the odd hole $H_N$ with complete outside vertex $u$ makes each vertex of $H_M$ complete to $H_N$. Applying the same lemma in $\bar{G}$ reverses the roles of hole and antihole and shows that each vertex of $H_N$ is anticomplete to $H_M$, contradiction. Therefore $G[N]$ has no odd hole, and by [L3] it must contain an odd antihole. Symmetrically, $G[M]$ contains an odd hole.

Take the odd antihole $H_N\subseteq G[N]$ and the odd hole $H_M\subseteq G[M]$ from step 2.1, with lengths $n$ and $m$. By [L2], each vertex of $H_N$ has at least $(m+1)\mathrm{/}2$ nonneighbors in $H_M$. Applying [L2] in the complement graph, where $H_N$ becomes an odd hole and $u$ is anticomplete to it, shows that each vertex of $H_M$ has at least $(n+1)\mathrm{/}2$ neighbors in $H_N$. Hence the number of cross nonedges is at least $n(m+1)\mathrm{/}2$ and the number of cross edges is at least $m(n+1)\mathrm{/}2$. Their sum is at least $(2mn+m+n)\mathrm{/}2>mn$, impossible. This contradiction proves that at least one of $G[N]$ and $G[M]$ is perfect.