Jérôme, it seems to me that your initial confusion may be due to (unfamiliarity with) the different notions of Turing reduction and Karp reduction (at least, this was the case for me when I incurred into that very same confusion years ago; judging from your question and comments I can perceive my own flawed objections I had at that time).
What you (seem to) suggest in your comment, i.e. to invoke a solver for the $\mathbf{NP}$ problem and to invert its answer in order to solve the corresponding $\mathbf{coNP}$ problem, is a Turing reduction from Non-Hamiltonian Graph to Hamiltonian Graph: in order to solve the former problem we just give the same input graph to a subroutine which is able to solve the latter problem, and return YES if and only if the subroutine returned NO. Doing so does not remotely constitute a proof that Non-Hamiltonian Graph and Hamiltonian Graph have the same complexity, nor that $\mathbf{coNP} \subseteq \mathbf{NP}$, it is just an instantiation of the fact that $\mathbf{coNP} \subseteq \mathbf{P^{NP}}$, i.e. that you can efficiently solve any $\mathbf{coNP}$ problem using an oracle for any $\mathbf{NP}$-complete problem.
Now try to do the same using a Karp reduction instead of a Turing reduction. This is going to be much more uncomfortable: you cannot invert the answer of the subroutine, nor you can perform any manipulation of such answer, you can only return it as it is. The mind-boggling difficulty of proving (or disproving) $\mathbf{NP} = \mathbf{coNP}$ precisely resides here.
Of course you cannot give the same input graph to the subroutine, as the returned answer would be flat wrong. You have to smartly manipulate the input graph, modifying it in some clever way, and then feed the subroutine with such modified graph, so that the subroutine (which in your suggestion is a solver for Hamiltonian Graph, but in principle can be a solver for any $\mathbf{NP}$-complete problem) returns YES if and only if the original input graph is Non-Hamiltonian. See how dramatically clever such manipulation would have to be: it would mean that any Non-Hamiltonian graph given in input is transformed into a Hamiltonian graph (more generally into a YES instance of any $\mathbf{NP}$-complete problem, e.g. into a satisfiable CNF formula), and vice-versa! Proving $\mathbf{NP} = \mathbf{coNP}$ is equivalent to furnish such a fantastic mapping, disproving it is equivalent to proving that no such mapping can exist.