If G has order $n \ge 3$ and for all pairs of distinct vertices $x$ and $y$ that are not adjacent, $deg(x)+deg(y) \ge n$ then the graph is Hamiltonian.
Here is the beginning of the proof:
We proceed by contradiction. Suppose there exists a non-Hamiltonian graph on $n$ vertices with the Ore property. Let G be a "maximal" such a graph, i.e. if we add a single edge to G, it becomes Hamiltonian.... the proof continues but I am wondering why exactly must G be connected?
The graph $G$ is assumed "maximal" in the sense that any extra edge forces the existence of a Hamiltonian cycle. If $G$ would consist of more than one component (and then be disconnected), we could add an edge going between two of these components. Then those two components, now connected, couldn't possibly have a Hamiltonian cycle through them, as it would start in one component and have to end in the other one (there's no edge back, so to say).
So adding that one edge would not force a Hamiltonian cycle. But then $G$ wouldn't be maximal, and there we have our contradiction.