Let $x \in X \setminus \bigcap_{j \in J} E_j$. Then $x \in X$ and $x \not\in \bigcap_{j \in J} E_j$. Since $x$ does not belong to the intersection, there is some $i \in J$ such that $x \not\in E_i$ and therefore $x \in X \setminus E_i$. Since $X \setminus E_i$ is part of the union, we have that $x \in \bigcup_{j \in J} (X \setminus E_j)$.
Conversely, let $x \in \bigcup_{j \in J} (X \setminus E_j)$. Then $x \in X \setminus E_i$ for some $i \in J$. That is, $x \in X$ and $x \not\in E_i$. Since $x \not\in E_i$, then also $x \not\in \bigcap_{j \in J} E_j$, and therefore $x \in X \setminus \bigcap_{j \in J} E_j$. $\square$
