Let $G=(V,E)$ be a simple graph with $E \subseteq P_2(V)$ and define $\overline{G}=(V,\overline{E})$ as the complement graph. Show that if $G$ is not connected, then $\overline{G}$ is connected.
I did the following: suppose that $G$ is not connected, then there exist $u,v \in V$; $u \neq v$ such that $u$ and $v$ are not connected. Let $x,y \in V$; if $x,y$ are not connected in $G$, then $\{x,y\} \notin E$, so $\{x,y\} \in \overline{E}$ and $x,y$ are connected in $\overline{G}$.That implies the validity for $|V|=2$. Now assuming that $|V|>2$,if $x,y$ are connected in $G$ and $x\neq u,v$ and $y \neq u,v$, then neither $x$ nor $y$ are mutually connected to $u$ and $v$ in $G$ as this'd imply that $u,v$ were connected in $G$. So there is at least an edge $e \in \overline{E}$ such that $e=\{x,u/v\} \in \overline{E}$ and an edge $e'$ such that $e=\{y,u/v\}$. But now, $x,y$ are connected in $\overline{G}$ as there exists a path $P$ between $x,y$ that passes through $u$ or $v$. Now if WLOG, $x=u \lor x=v$, there would be a path $P$ between $x$ and $y$ that uses $e$. So $\overline{G}$ must be connected. Is this a valid answer?
If the graph $G$ is finite then you can prove this easily. Take any two arbitrary vertices $x,y\in V(G)$. If $x$ and $y$ are not connected in $G$ then by definition of $\overline{G}$, $x$ and $y$ are connected in $\overline{G}$. If $x$ and $y$ are connected in $G$ then take a another vertices $v\in V(G)$ which is not connected to both $x$ and $y$ in $G$. Therefore in $\overline{G}$, $x$ and $v$ connected and $v$ and $y$ connected which means $x$ and $y$ are connected in $\overline{G}$. The existence of $v$ is due to the non-connectivity of $G$.