Every nontrivial $P_4$-free graph is disconnected or has disconnected complement

By [L1], if $G$ is disconnected then the first alternative of the Statement already holds. Assume instead that $G$ is connected.

Suppose for contradiction that $G$ is also anticonnected. Then [L2] makes $\bar{G}$ connected. Choose a vertex $v\in V(G)$. Because $G$ is connected and nontrivial, [F1] gives a neighbour $x$ of $v$. Because $\bar{G}$ is connected, [F1] also gives a vertex $y$ nonadjacent to $v$. Among all pairs $(a,b)$ with $va\in E(G)$ and $vb\mathrm{\notin }E(G)$, choose one for which the distance from $a$ to $b$ in $G$ is minimal. Let $a=u_0,u_1,\dots ,u_m=b$ be a shortest $a$-$b$ path in $G$.

Since $a$ and $b$ are nonadjacent, one has $m\ge 2$. A shortest path is induced, so $u_0{u}_2\mathrm{\notin }E(G)$ and no nonconsecutive pair among $u_0,u_1,\dots ,u_m$ is adjacent. If some $u_i$ with $1\le i<m$ were adjacent to $v$, then $(u_i,b)$ would be another neighbour/nonneighbour pair for $v$ whose distance is $m-i<m$, contradicting the choice in step 1.2. Hence $v{u}_1,v{u}_2\mathrm{\notin }E(G)$.

The four vertices $v,u_0,u_1,u_2$ therefore induce a path: the edges are $v{u}_0$, $u_0{u}_1$, and $u_1{u}_2$, while the nonedges are $v{u}_1$, $v{u}_2$, and $u_0{u}_2$. This is an induced copy of $P_4$, contradicting [L3].

So a nontrivial $P_4$-free graph cannot be both connected and anticonnected. Since step 1.1 reduced to the connected case, [L2] shows that whenever $G$ is connected, $\bar{G}$ must be disconnected.