Prove that if $ \ A\cup B \subseteq C \cup D,\ A \cap B =$ ∅ $\land \ C \subseteq A \implies B \subseteq D.$ Question:
Prove that if $ \ A\cup B \subseteq C \cup D,\  A \cap B =$ ∅ $\land \ C \subseteq A \implies B \subseteq D$.
My attempt:
Let $ \ x\in B \implies x \in A \cup B \implies x \in C \cup D \because A\cup B \subseteq C \cup D$.
Now, $ x \in C \lor x\in D$. If $\  x \in C \implies x \in  A \because C \subseteq A$. But that's not possible $\because x \notin A \cap B$, in particular $ x \notin A$. So we must have $ x \in D$.
I found this proof a little challenging. Not quite sure if this is the correct way to prove it. Is my logic correct?

It is correct. Well done.
Perhaps a bit of suggested modification in the "But that's not possible part" to make it clearer.
Since we started with $x \in B$, if $x \in A$, then $x \in A \cap B$ which is not possible, as $A \cap B = \phi$.
