I tried to solve this, but I don't think it is right. I suppose that $R$ is not reflexive and $R \not\subset R^*$ or $R^*$ is not reflexive.
Case 1: $R$ is not reflexive and $R \not\subset R^*$
We have that $R$ is not reflexive then $(x,x) \not\in R$, then $(x,x) \in R^c$. And $R \not\subset R^*$ then $R^c \subset R$, since $(x,x) \in R^c$ then $(x,x) \in R^*$. Thus $R^*$ is reflexive.
Let $I_X = \{ (a,a) \in X^2 \mid a \in X \}$ be the identity relation on $X$, and let $R$ be any relation on $X$. Then $R^* = R \cup I_X$ is reflexive and contains $R$. At this point, you should try to prove the claim. If you get stuck, then you can read the rest of this post for the solution.
Certainly, $R \subset R \cup I_X = R^*$, so $R$ is contained in $R^*$. To show reflexivity, let $x \in X$ be arbitrary. Then $(x,x) \in I_X \subset R \cup I_X = R^*$. Thus $R^*$ is reflexive.
Note that proof just given actually shows a stronger result. Whenever $R,S$ are two relations on a set $X$ and at least one of $R,S$ is reflexive, then the union $R \cup S$ is also reflexive.