 Solution

For every $a$, both $aRa$ and $aSa$ hold, so the intersection of equivalence relations is reflexive. If $a(R\cap S)b$, symmetry of both component binary relations gives $bRa$ and $bSa$, hence $b(R\cap S)a$. Finally, if $a(R\cap S)b$ and $b(R\cap S)c$, transitivity in each component gives $aRc$ and $aSc$, hence $a(R\cap S)c$. These are precisely the three conditions for an equivalence relation.