Understanding the derivation of Gibbs Free Energy In the equation: $$ΔG=ΔH−TΔS$$

I am having trouble understanding my professor's derivation of it from 
$$ΔS_{\mathrm{univ}}=ΔS_{\mathrm{sys}}+ΔS_{\mathrm{surroundings}}$$

where he says $$ΔS_{\mathrm{surroundings}}= \frac{-ΔH_{\mathrm{sys}}}{T}$$
which I have learned is only true for a reversible process. 
How is it possible that this heat can be reversed? And why can't we say that the change of entropy for the system is also equal to the equation. I understand that it would mean the reaction is at equilibrium, but would it not be since heat is reversible? 
$$ΔS_{\mathrm{sys}}= \frac{ΔH_{\mathrm{sys}}}{T}$$
The surroundings are treated as a constant volume reservoir, so the first law of thermodynamics,
$$
\mathrm{d}U_\mathrm{surroundings} 
  = \mathrm{d}q_\mathrm{surroundings} 
  + \mathrm{d}w_\mathrm{surroundings},
$$
yields 
$$
\mathrm{d}U_\mathrm{surroundings} 
  = \mathrm{d}q_\mathrm{surroundings}
  \text{, or }
\Delta U_\mathrm{surroundings} 
  = q_\mathrm{surroundings}.
$$ 
Notice that $q_\mathrm{surroundings}$ is equal to a state function, so it behaves as one, that is, it is independent whether the process is reversible or not. If the process is reversible
$$
q_\mathrm{rev,surroundings} 
  = \Delta H_\mathrm{surroundings},
  \tag1\label{eq:rev}$$
if it is not reversible 
$$
q_\mathrm{surroundings} 
  = \Delta H_\mathrm{surroundings},
  \tag2\label{eq:irrev}$$
from \eqref{eq:rev} and \eqref{eq:irrev} follows
$$q_\mathrm{surroundings} = q_\mathrm{rev,surroundings}.$$
Therefore 
$$
\Delta S_\mathrm{surroundings} 
  = \frac{q_\mathrm{rev,surroundings}}{T},
$$ 
in particular 
$$q_\mathrm{rev,surroundings} = -q_\mathrm{sys},$$ 
then 
$$\Delta S_\mathrm{surroundings} = \frac{-q_\mathrm{sys}}{T}.$$
If the process occurs at constant pressure 
$$q_\mathrm{sys} = \Delta H_\mathrm{sys},$$ 
therefore 
$$\Delta S_\mathrm{surroundings}  = \frac{-\Delta H_\mathrm{sys}}{T}.$$