To prove that $B\subset A\rightarrow A=A\cup B$ is it necessary to consider the cases $A\cup B=\emptyset$ and $A\cup B\neq \emptyset$? For what I know, to prove $B\subset A\rightarrow A=A\cup B$ it suffices to prove that $A\subset (A\cup B)$ and $(A\cup B)\subset A$.
Prove that $A\subset (A\cup B)$ is trivial, because $A\cup B=\{x:x\in A\vee x\in B\}$.
My teacher in proving that $(A\cup B)\subset A$ separated the proof into two parts: $A\cup B=\emptyset$ and $A\cup B\neq \emptyset$. But I didn't understand why the need to separate this proof in two cases since
$x\in (A\cup B)\to x\in A \vee x\in B$
We know, by hypothesis, that $B\subset A$, so we can conclude that $(x\in A\vee x\in B)\leftrightarrow(x\in A)$ is tautology(using truth table). So saying $x\in A\vee x\in B$ is equivalent to saying $x\in A$ which implies
$x\in (A\cup B)\to x\in A \vee x\in B\equiv x\in A\Rightarrow (A\cup B)\subset A$ (note: the implication $p\to q$ is true whenever $p$ is false).
In my opinion in the proof that $(A\cup B)\subset A$ there was no need to consider two cases, but as my teacher has great knowledge about set theory I feel that I am making a mistake by not considering the two cases. That is why I ask for help in this doubt.
Obs.: If I used some logical symbol wrongly let me know, please, since I still know few things about propositional logic.

You are right, and I see no errors. We can also observe that  $A\subset B\iff \forall x\;(x\in B\implies x\in A).$ Therefore if $B\subset A$ then for all $x$ we have  $$x\in A\cup B)\implies  (x\in A\lor x\in B)\implies$$ $$\implies (x\in A\lor x\in A)\implies(x\in A)\implies$$ $$\implies(x\in A \lor x\in B)\implies(x\in A\cup B).$$
From this we infer that for all $x$ we have $x\in A\cup B\iff x\in A$.
This is in my style, which is almost never brief.
