let $X$ and $Y$ be elements of your set. you have to show that
$(g~ R~ X) \wedge (X~ R~ Y) \Rightarrow (g ~R ~Y)$
then only element $X \ni (g~ R~ X)=True$ is $X=g$
so you are reduced to determining if
$(g~ R~ g) \wedge (g~ R~ g) \Rightarrow (g~ R ~g)$ which is trivially true if reflexion hold which in this case does.
The case for $c$ is identical.