Let p,r,q be sets. Prove that if p is a subset of r and r is a proper subset of q, then p is a proper subset of q
Consider an element $\displaystyle x \in p$.
As $\displaystyle p \subset r$ it follows by definition of a subset that $\displaystyle x \in r$.
But as $\displaystyle r \subset q$, it follows that if $\displaystyle x \in r$ then $\displaystyle x \in q$.
So, given that $\displaystyle x \in p$ we have shown that $\displaystyle x \in q$ as well.
That is all you need to do to show that $\displaystyle p \subset q$.
But you still need to prove that p is a proper subset of q! That is, that there exist a least one member of q that is not a member of p. To do that use the fact that r is a proper subset of q. Notice that the problem does NOT say that p is a proper subset of r!