Hamiltonian split into Mass term and Decay Width I have encountered the following procedure several times now, and none of the sources ever explain the physical reason behind it:

The Hamiltonian $H$ is split into $M$ and $\Gamma$.
WHY?
Where does this come from?
Is there a rigourous proof?
 A: I'll give you a draft of the answer to put you on the right track. You should then be able to fill the hole and complete the details.
I think you just need to take the non-relativistic limit of a field theory where instead of the tree-level propagator you use the full 1PI propagator $\frac{1}{p^2-m^2-\Pi(p^2)}$
 where $\Pi(\pi^2)$ is the self-energy. If the particle was stable then $m$ is the mass that shows up as a simple pole, $\Pi(m^2)=0$. But, if the particle is unstable the pole is actually off the real axis and move in the complex plane acquiring an imaginary part. This is actually guarantee by the optical theorem. This imaginary part comes basically by the self-energy around the pole location. Since in the non-relativistic limit you are expanding around that pole, this residual imaginary part from $\Pi$ is (proprotional to)  what you have called  $\Gamma$.
A: Once you have open systems you loose unitary time evolution in general and hence Hermitian Hamiltonians. You can see this by taking the full system+environment and then tracing over the degrees of freedom of the environment. 
This can be made rigorous with several assumptions - Lindblad equation.
In your case it is more of an effective description for a simple two level system,. You can easily see that the time evolution defined by this generator leads to decay of excited states of the Hamiltonian. But this can too be derived under certain assumptions and approximations, see for instance arXiv:1207.4877
