This is Velleman's exercise 3.5.4:
Suppose $A \cap C \subseteq B \cap C$ and $A \cup C\subseteq B \cup C$. Prove that $A \subseteq B$
This is the proof given by the book (which I understand completely):
Proof. Suppose x ∈ A. We now consider two cases:
Case 1. $x \in C$. Then $x ∈ A \cap C$, so since $A \cap C\subseteq B \cap C, x \in B \cap C$, and therefore $x \in B$.
Case 2. $x \notin C$. Since $x \in A, x \in A \cup C$, so since $A \cup C\subseteq B \cup C$, $x \in B \cup C$. But $x \notin C$, so we must have $x \in B$.
Thus, $x \in B$, and since x was arbitrary, $A \subseteq B$.
I was wondering if one could write a proof like this one in below:
Proof. Let x be an arbitrary element of A. Then by $A \cup C \subseteq B \cup C$, we have either $x \in B$ or $x \in C$. Now we consider these two cases:
Case 1. x is an element of B.
Case 2. x is an element of C.
Since in one of the cases $x \in B$ and since x was arbitrary, $A \subseteq B$.