I'm not very clever, and thinking gives me a headache. Let's try to solve this thing with a minimum of cleverness and hard thinking. A little brute force is OK.
Let's try for a proof by contradiction, that will always work if anything does. So we assume a disconnected graph $G$ with disconnected complement $\overline G.$
Then there are two vertices $a,b$ which can't be connected by a path in $G,$ and there are two vertices $c,d$ which can't be connected by a path in $\overline G.$
Let $H$ be the subgraph of $G$ induced by $\{a,b,c,d\},$ a graph with $2,3,$ or $4$ vertices depending on how $\{a,b\}$ and $\{c,d\}$ overlap. (Maybe I could eliminate one of those cases if I thought about it. Nuts to that.) If two vertices can't be connected by a path in the big graph, then they can't be connected by a path in the subgraph, right? So $H$ and $\overline H$ are still disconnected.
Great, that means I only have to check disconnected graphs with $2,3,$ or $4$ vertices. I can do that with brute force!
Allowing myself a tiny bit of cleverness to save work: If my disconnected graph $H$ has a disconnected complement, and if I can add edges to $H$ without making it connected, then the complement (which is losing edges) will still be disconnected! So I only have to test maximal disconnected graphs!! With just a little bit of thought, I see that a maximal disconnected graph is the sum of two complete graphs. Up to $4$ vertices, the only graphs I have to check are: $K_1+K_1,\ K_1+K_2,\ K_1+K_3,\ K_2+K_2.$ The complements of those graphs are $K_{1,1},\ K_{1,2},\ K_{1,3},\ K_{2,2}.$ The complements are all connected! That does it!!
Wait a minute, that's looking suspiciously similar to the proofs in the other answers.