Why does the derivation Gibbs-Helmholtz equation hold when the enthalpy is temperature-dependant
The derivation of the van't Hoff equation does not require that you specify the T-dependence of H. It arises from computing the derivative (limiting slope) of $G/T$ so you need to consider a polynomial expansion of $G/T$ to linear terms in $T$ around the temperature of interest (that is, an infinitely  small interval dT). At the risk of redundancy: 

$$\begin{align}
 (d(G/T))_p &amp;= ((1/T)dG-G(dT/T^2))_p \\
&amp;= ((1/T)(VdP-SdT)-(H-TS)(dT/T^2))_p \\ 
&amp;= (-(S/T)dT-(H/T^2-S/T)dT)_p \\
&amp;= (-(H/T^2)dT)_p 
\end{align}
$$

The answer is therefore found in differential Calculus.

Another comment explains that, if 

$G(T)=H(T) -TS(T)=H_0  +C_p \Delta T -TS(T)$ 

where $H(T)$ is expressed as a linear polynomial in T ($C_p$ and $p$ assumed constant) then  

$dG=C_p dT -SdT-TdS$

Since for a process at constant pressure,  $dq = C_p dT=TdS$ (since $C_p = (dH/dT)_p$ and $dq_p= dH$; also for a reversible process $dq= TdS$), it follows that 

$\left(\frac{ \partial G}{\partial T} \right)_p=-S$

and the van't Hoff expression provided in the linked answer also follows.