Proof. Suppose $R$ and $S$ are both equivalence relations on a set $A$. In what follows we will show that $R$ and $S$ both being equivalence relations on the set $A$ implies that $R\cap S$ is also an equivalence relation.
It is immediately apparent that since $R$ and $S$ are equivalence relations then
$\begin{align*} x\in A \Rightarrow (xRx)\land(xSx) \Rightarrow (\langle x,x\rangle\in R) \land (\langle x,x\rangle\in S)\Rightarrow\langle x, x\rangle\in R\cap S.\end{align*}$
Thus $x\in A\Rightarrow x(R\cap S)x$ therefore $R\cap S$ is reflexive.
Now suppose $\langle x,y \rangle\in R\cap S$ then $(\langle x,y \rangle\in R) \land (\langle x,y \rangle\in S)$ by definition of the intersection of two sets. But both $R$ and $S$ are symmetric so it must be that $(\langle y,x \rangle\in R) \land (\langle y,x \rangle\in S)$ which implies $\langle y,x \rangle\in R\cap S$. Thus we have shown $x(R\cap S)y\Rightarrow y(R\cap S)x$ therefore $R\cap S$ is symmetric.
Finally, consider $(\langle x,y \rangle\in R\cap S)\land(\langle y,z \rangle\in R\cap S)$. Then since $(\langle x,y \rangle\in R)\land\langle (y,z \rangle\in R)\Rightarrow\langle x,z \rangle\in R$, since $R$ is transitive, and since $(\langle x,y \rangle\in S)\land\langle (y,z \rangle\in S)\Rightarrow\langle x,z \rangle\in S$, since $S$ is transitive, therefore $\langle x,z\rangle\in R$ and $\langle x,z\rangle\in S$ so by definition of the intersection of two sets $\langle x,z\rangle\in R\cap S$. Thus we have shown $(\langle x,y \rangle\in R\cap S)\land(\langle y,z \rangle\in R\cap S)\Rightarrow\langle x,z\rangle\in R\cap S$.
It follows that $R\cap S$ is an equivalence relation since it is reflexive, symmetric and transitive. $\Box$