In a connected and anticonnected graph, the union of two proper modules that meet is again a proper module

The set $M\cup N$ is a module of $G$ by [L1], since $M\cap N\mathrm{\ne }\mathrm{\varnothing }$. Suppose for contradiction that $M\cup N=V(G)$.

The set $M$ is nonempty, since it contains a vertex of $M\cap N$, and $M\mathrm{\ne }V(G)$ because $M$ is proper.

If $M\setminus N$ were empty then $M\subseteq N$, so $N=M\cup N=V(G)$ by step 1.1, which is false because $N$ is proper; hence $M\setminus N\mathrm{\ne }\mathrm{\varnothing }$, and $V(G)\setminus M=N\setminus M$, again by step 1.1.

Since $G$ is connected and $M$ is a module with $\mathrm{\varnothing }\mathrm{\ne }M\mathrm{\ne }V(G)$, some vertex $v\in V(G)\setminus M$ is complete to $M$.

The set $M$ is a module of $\bar{G}$ as well, and $\bar{G}$ is connected with the same vertex set as $G$, so some vertex $v^{\mathrm{\prime }}\in V(G)\setminus M$ is complete to $M$ in $\bar{G}$; that is, $v^{\mathrm{\prime }}$ is adjacent in $G$ to no vertex of $M$.

Choose $x\in M\setminus N$. Then $x\mathrm{\notin }N$, while $v$ and $v^{\mathrm{\prime }}$ both lie in $V(G)\setminus M=N\setminus M\subseteq N$, so [L4] applied to the module $N$ gives that $xv\in E(G)$ if and only if $x{v}^{\mathrm{\prime }}\in E(G)$.

But $v$ is complete to $M$ and $x\in M$, so $xv\in E(G)$, while $v^{\mathrm{\prime }}$ is adjacent to no vertex of $M$, so $x{v}^{\mathrm{\prime }}\mathrm{\notin }E(G)$; this contradicts step 3.1. Hence $M\cup N\mathrm{\ne }V(G)$, and being a module it is a proper module.