Proof that if $A$ is a subset of $B$ and $C$ is a subset of $D$, then $A \setminus D$ is a subset of $B \setminus C$. Started my proof by assuming that $A \subset B$ and $C \subset D$. Then I proceeded by supposing that $x \in  A \setminus D$. So $x \in  A$ and $x \notin D$. Is it valid to conclude that $x \notin C$ since  $C \subset D$? I'm assuming it is always true but I'm not sure if it is a valid argument I can use for formal proofs.
Aside from that, I'm also lost as to how I should continue my proof. For now I continued with my proof mentioned above and added &quot;since $A \subset B$, then $x \in B$. Also, $x \notin C$ so it follows that $x \in B \setminus C$. Therefore,  $A \setminus D \subset B \setminus C$. &quot; Although, I'm not really sure if this is a valid argument. For instance, I'm having doubts with $x \in B$ and $x \notin C$ since $x$ is also not an element of $D$. Therefore, there's a possibility that $x \in B \setminus D$ (??). Any help with regards as to how I should properly and formally prove this will be a great help.
Edit: Thanks a lot to everyone who verified my approach!
You're correct stating that if $x\not\in D$, since $C\subset D$ then $x\not\in C$. If it isn't clear to you, look at the contraposition (if $x\in C$, since $C\subset D$ then $x\in D$).
In the second part, you proved that for every element $x\in A\setminus D$, you have $x\in B\setminus C$, therefore $A\setminus D\subset B\setminus C$. If $x\in B$ and $x$ is not an element of $D$ then automatically $x\in B\setminus D$.
Your argument is right. In symbols, you're essentially doing $$A\subset B\quad\text{and}\quad D^c\subset C^c\implies A\cap D^c\subset B\cap C^c$$
using the fact that $M\subset N\iff N^c\subset M^c$, which is right.