Fix $i \in J$ and $x \in \bigcap_{j \in J} E_j$. By definition of set intersection, $x \in E_j$ for every $j \in J$ and thus also for $i$. This proves (1).

Suppose then that $X \subseteq E_j$ for every $j \in J$. If $X$ is empty, then the claim is a consequence of R7: Empty set is subset of every set, so we may assume that $X$ is nonempty and fix $x \in X$. Since $X$ is contained in $E_j$ for each $j \in J$, then $x \in E_j$ for every $j \in J$ and thus, by definition of set intersection, $x \in \bigcap_{j \in J} E_j$. This concludes the proof. $\square$