Given a simple graph and its complement, prove that either of them is always connected. I was tasked to prove that when given 2 graphs $G$ and $\bar{G}$ (complement), at least one of them is a always a connected graph.
Well, I always post my attempt at solution, but here I'm totally stuck. I tried to do raw algebraic manipulations with # of components, circuit ranks, etc, but to no avail.
So I really hope someone could give me a hint on how to approach this problem.

Let $G$ be a simple disconnected graph
$\Longrightarrow\ \exists$ atleast $2$ vertices, say $u$ and $v$ such that there does not exist a path between $u$ and $v$.
$\Longrightarrow$ All vertices of $G$ are not adjacent to both $u$ and $v$ (Why?). $u$ and $v$ are obviously not adjacent.
$\Longrightarrow$ In $\bar{G}$ all the vertices are adjacent to either $u$ or $v$ or both and $u$ is adjacent to $v$.
$\Longrightarrow$ Take any two vertices in $\bar{G}$, say $a$ and $b$, we see that $a$ can be adjacent to either $u$ or $v$ or both. Same holds for $b$. Since, $u$ and $v$ are adjacent, there must exist a path through either $u$ or $v$ or both that connects $a$ and $b$.
$\Longrightarrow$ Hence, $\bar{G}$ is connected.
Same holds for $G$, if $\bar{G}$ is disconnected. Thus, at least one of $G$ and $\bar{G}$ must be connected.
