$Proof$. Suppose $x\in Dom(S)$ and $x\notin Dom(R)$. Then there is some $y\in A$ such that $xSy$. Since $S$ is the transitive closure of $R$, there exist some set of pairs in $R$ such that $x$ and $y$ can be connected. But $x\notin Dom(R)$, so this connection cannot be made. Therefore it must be that $x\in Dom(R)$.
Im not 100% confident that this proof is valid. Is the wording used ok? If $S$ is the transitive closure of $R$ and $xSy$, then if you understand what transitive closure means, you would know that there are some $xRz$ and $zRb$ and $bRy$ to make $xSy$, but I feel like the way I explain it in the proof would be considered bad.
