Let $G=(X,U)$ be a simple graph, with $n$ vertices, not Hamiltonian and such that, for any distinct and non-adjacent $x_i$ and $x_j$ vertices, $G+\{x_i,x_j\} $ is Hamiltonian.
I need to show that
(i) $ G $ is connected.
Suppose, absurdly, that $G$ is disconnected, so we get at least two connected components in $G$. Note that any vertex of a connected component $S_1$ is not adjacent to vertices of the other connected component $S_2$. So if we add an edge that connects a vertex of $S_1$ to a vertex of $S_2$ we have that this edge will be a bridge, and we know that for any two non-adjacent vertices of $G$ we have to $G+\{x_i,x_j\}$ is a Hamiltonian graph, i.e., has a Hamiltonian cycle. Therefore, we have that $G+\{x_i,x_j\}$ has no bridges. In case the two non-adjacent vertices are of the same connected component, we would never have a generator cycle. Then we conclude that $G$ must be a connected graph.
(ii) If $x_i$ and $x_j$ are distinct and not adjacent, then $x_i$ and $x_j$ are consecutive vertices of any Hamiltonian cycle of $G+\{x_i,x_j\}$.
(iii) If $x$ is a vertex of degree 1 of $G$ then $G-x$ is isomorphic to $K_{n-1}$.
I'm not sure if the first proof is it correct, can anyone help me to prove the second and third statement and verify the first proof?