Spontaneity of an exothermic reaction when temperature is decreased
The key is $\Delta G = -RT \ln K$. 

The $T$ specifically is causing your reasoning to be flawed, as it complicates the link between temperature dependence of the equilibrium constant and $\Delta G$. 

$-RT \ln K = \Delta H-T\Delta S$

$-R \ln K = \frac{\Delta H}{T}-\Delta S$

In this form you can see that (at least to the extent that $\Delta H$ and $\Delta S$ are temperature independent), the sign of $\Delta H$ (meaning whether the reaction is endothermic or exothermic) determines whether or not temperature change favors the forward or the reverse reaction. You can also see that the sign of $\Delta S$ is irrelavent. 

For an exothermic reaction, $\Delta H$ is negative.  If $T$ decreases, $\frac{\Delta H}{T}$ and becomes a larger magnitude negative number, $-R\ln K$ decreases, $R\ln K$ increases therefore $K$ increases and the reaction shifts forward.