Let $A = A\cup B$ be true, it suficces that $B=\emptyset$, but, as we observe ,we are trying to prove $A\subseteq A\cup B$ for arbitrary sets $A$ and $B$. Should $B=\emptyset$, then we are stating that any arbitrary sets is empty, which contradicts the axiom of existence: "There exists a set with no elements".
Since we know $A=A\cup B$ is not true for all sets $A$ and $B, A\subseteq A\cup B$ must be true. That being so, $A\subseteq A\cup B$ is trivially true.
Is this proof right?
You are looking at the sign $\subseteq$ in a very inefficient way, and making things more complicated than necessary!
Yes for inequalities like $X\le Y$, it means $X=Y$ or $X<Y$. Similary, $A\subseteq B$ means "$A=B$" or "$A$ is a proper subset of $B$", i.e., $A\subsetneq B$. But you don't need this!
Simply go with the definition: $P\subseteq Q$ means any element in $P$ is an element in $Q$.
So, is any element in $A$ an element in $A\cup B$? Of course, being in $A\cup B$ means either in $A$ or either in $B$.
QED.