Prove that $\text{Dom } (S\circ R) ⊆ \text{Dom }R $ Let $R$ be a relation from $A$ to $B$ and $S$ be a relation from $B$ to $C$.
Suppose, $x \in \text{Dom }(S\circ R)$. Then, it follows that there $\exists y \in C$ such that $(x,y) \in S\circ R $. Since $S\circ R$ is a subset of $A\times C$ and by the definition of $\text{Dom }(S\circ R)$. 
Thus there $\exists y \in C$ such that $(\exists r \in B)[(x,r)\in R$ and $(r,y)\in S ]$. 
Therefore, it follows that $x\in \text{Dom }R$, since the statement $(\exists r \in B)(x,r)\in R$ satisfies the requirement for membership in $\text{Dom }R$.
Is this proof correct? Thank you!

Your proof is correct.
Alternative: 
If $a\notin\text{Dom }R$ then no $b\in B$ exists with $aRb$. 
Consequently no $b\in B$ and $c\in C$ exist with $aRb\wedge bSc$. 
This allows the conclusion that $a\notin\text{Dom }(S\circ R)$.
