Hint: Use the fact that $R$ and $S$ are EQUIVALENCE relations on THE SAME set, and hence both must be reflexive, symmetric, and transitive on that set.
Then use the definition of set intersection: $R\cap S$ is the set of all pairs of elements in the set such that $(x, y) \in R$ AND $(x, y) \in S$ or, put differently, $(x, y) \in R\cap S \iff (x, y)\in R$ and $(x, y) \in S$.
Try to figure out what elements must necessarily be in $R\cap S$ and check to see that they must then be in both $R$ and $S$.
Another approach would be to use an indirect proof with the hints above:
"Given $R$ and $S$ are equivalence relations on a set $A$, suppose for the sake of contradiction, that $R\cap S$ is NOT an equivalence relation...". If not an equivalence relation, then $R\cap S$ fails to be reflexive and/or fails to be symmetric, and/or fails to be transitive. If you can work towards a contradiction (that this assumption must contradict the fact that both $R$ and $S$ are equivalence relations), then you are done.