Let $E_j$ and $F_j$ each be a D11: Set for every $j \in J$ such that

(i)	\begin{equation} \forall \, j \in J : E_j \subseteq F_j \end{equation}

If the intersection $\bigcap_{j \in J} E_j$ is empty, the claim follows from R7: Empty set is subset of every set. Suppose thus that $\bigcap_{j \in J} E_j$ is not empty and fix $x \in \bigcap_{j \in J} E_j$. By definition of set intersection, now $x$ belongs to $E_j$ for every $j \in J$. Combining this with the initial assumptions, one then has \begin{equation} x \in E_j \subseteq F_j \end{equation} and thus $x \in F_j$ for every $j \in J$. Therefore, again by the definition of a set intersection, $x \in \bigcap_{j \in J} F_j$. Since $x \in \bigcap_{j \in J} E_j$ was arbitrary, we have the inclusion $\bigcap_{j \in J} E_j \subseteq \bigcap_{j \in J} F_j$. $\square$