Prove that if $A \cup B$ $\space\subseteq $ $\space C \cup D$, $A \cap B$=$\emptyset$ and $C\subseteq A$, then $B \subseteq D.$
I tried working around with this for a while and reached this conclusion, however I do not know if I am making any sense at all:
Let $y \in$ $A \cup B$ $\subseteq $ $ C \cup D$
Since $(y \in A$ or $y \in B\space)$ $\subseteq \space$ $( y \in C$ or $y \in D)$ and because $A \cap B$=$\emptyset$ which means that A and B are disjoint, $C\subseteq A$ implies that $B \nsubseteq C$ because A and B are disjoint. However since $A \cup B\space$ $\subseteq \space$ $ C \cup D$, $B$ MUST be a subset of $D$ as $B \nsubseteq C$.
Sorry if I restated the same thing a few times.