Let $\Omega$ be a set and $\mathscr{E}$ be a family of disjoint subsets of $\Omega$. Show that for any $A \in \mathscr{E}$ and any $B \in \sigma(\mathscr{E})$, we have either $A \subset B$ or $A \cap B = \varnothing$.
How should I start to proof this rigoursly? Should I give all the options possible?
Let $\mathcal{F} = \{B : \text{ for all } A\in \mathscr{E}, A\subseteq B \lor A\cap B = \emptyset \}$
Clearly $\mathscr{E}$ is contained in $\mathcal{F}$. Therefore if we show that $\mathcal{F}$ is a $\sigma$-algebra we get that $\mathscr{E}\subseteq\mathcal{F}$ which is what required.
Indeed $\emptyset,\Omega\in \mathcal{F}$. Also if $B\in \mathcal{F}$ then $B^c\in \mathcal{F}$ because $A\subseteq B \Rightarrow A\cap B^c=\emptyset$ and $A\cap B=\emptyset \Rightarrow A\subseteq B^c$.
Finally, let $B_1,B_2,...\in\mathcal{F}$ we claim that the union is in $\mathcal{F}$. Let $A\in\xi$, if there exists $i$ such that $A\subseteq B_i$ then $A\subseteq \bigcup B_i$. Otherwise $A\cap B_i=\emptyset$ for all $i$ and so $A\cap \bigcup B_i = \emptyset$. This completes the proof.