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 $\bigcup_{j \in J} E_j$ is empty, then the claim is a consequence of R7: Empty set is subset of every set. Suppose thus that $\bigcup_{j \in J} E_j$ is not empty and fix $x \in \bigcup_{j \in J} E_j$. By definition of set union, there is some index $i \in J$ such that $x \in E_i$. Since $E_i \subseteq F_i$, then $e \in F_i$ for every $e \in E_i$. In particular, $x \in F_i$ since $x \in E_i$. Thus, by definition of set union, $x \in \bigcup_{j \in J} F_j$. Since $x \in \bigcup_{j \in J} E_j$ was arbitrary, the proof is complete. $\square$