If $R$ is an equivalence relation, does $R^2$ too? I think that yes,

$I_A \subseteq R$
$R = R^{-1}$
$R^2 \subseteq R$

And now we can show.
Reflex: $I_A = I_A^2 \subseteq R \subseteq R^2$
A lil bit struggling with symm. And trans. Can you show me how will you prove it?

The key is that $R_1\subseteq R_2$ implies $R_1^2\subseteq R_2^2$. Together with $I_A^2=I_A$ and $(R^{-1})^2=(R^2)^{-1}$ the three properties for $ 2$ follow.
But actually the situation is much simpler: While transitivity (alone) of $R$ in fact just means $R^2\subseteq R$, transitivity plust reflexivity means that $$R^2=(R\cup I_A)^2=R^2\cup RI_A\cup I_AR\cup I_A^2=R^2\cup R\cup I_A=R. $$
Hence $R^2$ is an equivalence relation because it is the same relation as $R$. (We also not that we did not need symmetry to show $R^2=R$, only transitivity and reflexivity; hence $R^2=R$ holds as well for relations such as $\le$)
