First a little bit about transitive closure. Let $R$ be a binary relation on $A$. The transitive closure $\tau(R)$ of $R$ can be defined in several different ways. First $\tau(R)$ is the smallest transitive set containing $R$. It can be defined as $\tau(R) = \bigcup_{n \in \mathbb N} R^n$. One other property that can be used to define transitive closure is $(x, y) \in \tau(R)$ iff $\exists z_0, ..., z_n \in A$ such that $x = z_0$, $y = z_n$ and $(z_i, z_{i+1}) \in R$ for $i = 0, ..., n - 1$. This is easy to derive from $\tau(R) = \bigcup_{n \in \mathbb N} R^n$.
Back to the question. Assume $(x, y) \in \tau(R_1 \cap R_2)$. Then there are $x = z_0, z_1, ..., z_n = y \in A$ such that $(z_i, z_{i+1}) \in R_1 \cap R_2$. But then $(z_i, z_{i+1}) \in R_1$ and $(z_i, z_{i+1}) \in R_2$ for $i = 0, ..., n - 1$. Then $(x, y) \in \tau(R_1)$ and $(x, y) \in \tau(R_2)$ and so $(x, y) \in \tau(R_1) \cap \tau(R_2)$.