I wrote the following inductive proof, but I don't know if it is valid and if there are any points that I'm missing. I got a bit lost trying to prove the inductive step.
We prove this statement by induction on the number of vertices for $G$ and $\bar{G}$. Base case n = 1. G with one vertex and its complement are the same connected graph because a single vertex is connected to itself by a path with length zero. Then we establish the inductive hypothesis where the statement is true for $n = k$ vertices in $G$ and $\bar{G}$. We proceed to show that the statement is true for $n = k+1$ vertices in $G$ and $\bar{G}$. Let $v$ be the extra vertices in $k + 1$. According to the induction hypothesis, either $G$ or $\bar{G}$ with $k$ vertices are connected, so either $G - v$ or $\bar{G} - v$ is connected. Suppose for the sake of contradiction that both G and $\bar{G}$ are disconnected. If G - v is a connected graph, then there must not be a vertex in $G-v$ that connects to v in G as the graph would otherwise be connected. Since $\bar{G}$ is a complement of G, two vertices that are not connected in $G$ are connected in $\bar{G}$. Since every two vertices other than $v$ in $G$ are connected, in $\bar{G}$ every vertex in G-v must connects to $v$, which makes $\bar{G}$ a connected graph, which is a contradiction. Then if $G - v$ is a connected graph, G and $\bar{G}$ must not both be disconnected, so either $G$ or $\bar{G}$ is connected. Similar argument may be made for the case if \bar{G}-v is a connected graph, in this case either G or \bar{G} must be connected. Therefore, the statement is true for $n = k+1$. Therefore, we have proven that either $G$ or its complement $\bar{G}$ must be a connected graph.
ETA: I think see what you were trying to do now. In any event what you want to note is below:
HINT: Suppose $G-\{v\}$ is connected. Then on the one hand, if $v$ is incident to any edge in $G$, then $G$ is also connected. Then on the other hand, if $v$ is not incident to an edge in $G$, then for every other vertex $u \in G$, there is an edge from $u$ to $v$ in $\bar{G}$. So $\bar{G}$ must be connected [why is that].
And likewise, if $\bar{G}-\{v\}$ that is connected.
There is at least one confusing/incorrect statement in your writeup though: What do you mean "since every two vertices other than $v$ in $G$ are connected...". That's not true. You indeed want an edge in $\bar{G}$ between $v$ and each other vertex, but this follows because $v$ is not incident to any edges in $G$.