Prove that $\text{Id}_{\text{dom }\mathcal R}⊆\mathcal R^{-1}∘\mathcal R$ and $\text{Id}_{\text{rank }\mathcal R}⊆\mathcal R∘\mathcal R^{-1}$

Simply use the definitions and the usual strategy for proving that a set inclusion holds: To prove that $A \subseteq B$, show that for any $a \in A$ we also have that $a \in B$.

If you want to show

$$ \text{Id}_{\text{dom}\;\mathcal R}\subseteq\mathcal R^{-1}\circ\mathcal R $$

you must show that for any $(x,x) \in \text{Id}_{\text{dom}\;\mathcal R}$ we also have that $(x,x) \in \mathcal R^{-1}\circ\mathcal R$. Since $(x,x) \in \text{Id}_{\text{dom}\;\mathcal R}$, we know that $x \in \text{dom}\;\mathcal{R}$. This then implies that there exists a $y$ such that $(x,y) \in \mathcal{R}$. And therefore $(y,x) \in \mathcal{R}^{-1}$. The result now follows by the definition of relation composition. The other inclusion is proved in the same way.