Prove a relation for a set If $ R,S $ are relations on the set $ A $, where $ S $ is reflexive

$ S \subseteq R $ 
Prove that:
$ R $ is reflexive
How do I begin? How could a relation be a subset of another relation ? thanks
 A: A relation is a set of ordered pairs, i.e.a subset of $A^2$. So both $R$ an $S$ are subsets of $A^2$ and one could be a subset of the other. Take any $a\in A$. Since $S$ is reflexive we have that $aSa$, that is $\langle a,a \rangle\in S$. But $S\subseteq R$, so $\langle a,a \rangle\in R$, that is $aRa$. Since this holds for every $a\in A$, the relation $R$ is reflexive. 
Edit. Just an example to clarify the concepts. Say $A$ is the set of all real numbers, $S$ is the equality relation, $aSb$ iff by definition $a=b$, and finally let $R$ be the $\le$ relation, that is $aRb$ iff $a\le b$ (where as usual iff abbreviates if and only if). 
In this case $S=\{(x,y):x=y\}$, that is $S$ is (the graph of) the line $y=x$ (as a subset of the plane $A^2$), which is usually called the diagonal $\Delta$. At the same time, $R=\{(x,y):x\le y\}$ is the half-plane $H$ above (and to the left) of the diagonal, including the diagonal. 
Both relations contain $\Delta$, so both are reflexive. Also, $\Delta\subseteq H$, in other words $S\subseteq R$, sometimes this may even be written as $=\subseteq \le$. 
A: Hint:
$$(x,x) \in S \subset R $$
A: Saying that a relation $X$ on the set $A$ is reflexive means that $\Delta_A\subseteq X$, where
$$
\Delta_A=\{(x,x):x\in A\}
$$
Then
$$
\Delta_A\subseteq S
$$
and, since $S\subseteq R$, …
