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.

The reason you feel there is no reason to consider the two cases is because you feel that one case is trivial.   Indeed the conclusion from those premises is a vacuuously truth:
$$\begin{array}{lrcl}\because&(B\subseteq A)~,~ (A\cup B=\emptyset) &\vdash& \forall x ~\big((x\in A\cup B) \to (x\in A)\big)\\\hline\therefore&(B\subseteq A)~,~ (A\cup B=\emptyset)&\vdash&(A\cup B\subseteq A)\end{array}$$

However, not all proofs will have a trivial result from the empty case; for instance any proof involving an existance claim may well lead to an exception.   Your teacher just wants you to get into the habit of verifying a proof does not fail in the empty case.
