The first claim is clear since $E_i$ forms part of the union $\bigcup_{j \in J} E_j$ for each $i \in J$.

Let $E_j$ then be contained in $X$ for each $j \in J$. Let $x \in \bigcup_{j \in J} E_j$. Then there is an index $i \in J$ such that $x$ belongs to $E_i$. Since $E_i \subseteq X$, transitivity of subset relation implies that $x \in X$. Since $x \in \bigcup_{j \in J} E_j$ was arbitrary, this concludes the proof. $\square$