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.
Suppose $G$ is disconnected. We want to show that $\bar{G}$ is connected. So suppose $v$ and $w$ are vertices. If $vw$ is not an edge in $G$, then it is an edge in $\bar{G}$, and so we have a path from $v$ to $w$ in $\bar{G}$. On the other hand, if $vw$ is an edge in $G$, then this means $v$ and $w$ are in the same component of $G$. Since $G$ is disconnected, we can find a vertex $u$ in a different component, so that neither $uv$ nor $uw$ are edges of $G$. Then $vuw$ is a parth from $v$ to $w$ in $\bar{G}$.

This shows that any two vertices in $\bar{G}$ have a path (in fact a path of length one or two) between them in $\bar{G}$, so $\bar{G}$ is connected.
This is another answer, showing G¯ is connected in 3 cases by showing there exists an {a,b} path in each case. Let x,y be vertices in G. Let H be the component that holds the vertex x but not y.

C1: Suppose a and b are arbitrary vertices not in H. Then x~a and x~b in G¯ so there is a unique {a,b} path in G¯, namely {a,x,b}. (~ means adjacent)

C2: Suppose H holds one of the two vertices, a and b. Without loss of generality suppose H holds a. Then x~b and a~b in G¯. So there is a unique {a,b} path in G¯, namely {a,b}.

C3: Suppose H holds both the a and b vertices. Then y~a, y~b, y~x in G¯. So there is a unique {a,b} path in G¯, namely {a,y,b}.

Therefore there is a unique {a,b} path in G¯ for arbitrary a and b vertices in all 3 cases. Therefore G¯ is connected.