First of all, How can you assume that, $\Delta S$ is positive in any exothermic reaction? Secondly, the relation you have written $\Delta G = -RT\ln K_\text{eq}$ is not at all correct. The correct relation is $\Delta G = \Delta G^\circ + RT\ln Q_c$ (or $Q_p$ if the reactants are gaseous). Now at equilibrium $\Delta G= 0$ and $Q = K$. So, the correct relation is
$$\Delta G^\circ = -RT\ln K_\text{eq}$$
And, there are obviously differences between $\Delta G$ and $\Delta G^\circ$. So, the reasoning you have given is not totally correct.
The correct reasoning comes from Van't Hoff Equation, which states $$\frac{\mathrm d}{\mathrm dT}(\ln K_{\text{eq}}) = \frac{\Delta H}{RT^2}$$
Integrating both sides, we have
$$\int_{K_1}^{K_2}\mathrm d(\ln K_{\text{eq}} ) = \int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\,\mathrm dT$$
So, we have $$\ln\frac{K_2}{K_1}= \frac{\Delta H}{R}\left(\frac{1}{T_1} -\frac{1}{T_2}\right)$$ Thus we see that increasing the temperature from $T_1$ to $T_2$, the equilibrium constant changes. Here $\Delta H < 0$ (exothermic reaction) and $\displaystyle\frac{1}{T_1}- \frac{1}{T_2} > 0$ so, $\displaystyle \ln\left(\frac{K_2}{K_1}\right) < 0$, which means $\displaystyle {\frac{K_2}{K_1}< 1}$, which implies $K_2 < K_1$. So, the equilibrium constant decreases.
Note: Without any mathematical equation also, the same result can be deduced. In an exothermic reaction, the products are formed along with heat production, which is the definition of exothermic reaction. That means, temperature of the system will increase due to this heat production. Now if you increase the temperature, the system will obviously want to go in the direction where this temperature minimisation is possible, which is actually the backward reaction. So, the reaction shifts backwards, and thus the equilibrium constant decreases as now the product concentration has become lesser and reactant concentration has become larger.
