In search of contradiction, assume that $\bigcap_{j \in J} X_j \neq \emptyset$ and fix $x \in \bigcap_{j \in J} X_j$. By definition of set intersection, we have that $x \in X_j$ for every $j \in J$. By hypothesis, there now exists an index $i \in J$ such that $x \in X_i = \emptyset$. That is, $x \in \emptyset$, which is a contradiction and hence the equality $\bigcap_{j \in J} X_j = \emptyset$ must hold. $\square$