This is one of the problem I have been working from Velleman's How to Prove it
book:
Theorem: Suppose $A$, $B$, and $C$ are sets and $A \subseteq B \cup C$. Then either $A \subseteq B$ or $A \subseteq C$. Is the theorem correct?
Proof. Let $x$ be an arbitrary element of $A$. Since $A \subseteq B \cup C$, it follows that either $x \in B$ or $x \in C$.
Case $1.$ $x \in B$. Since $x$ was an arbitrary element of $A$, it follows that $\forall x \in A(x \in B)$, which means that $A \subseteq B$.
Case $2.$ $x \in C$. Similarly, since $x$ was an arbitrary element of $A$, we can conclude that $A \subseteq C$. Thus, either $A \subseteq B$ or $A \subseteq C$.
After reading the proof, I was pretty much convinced that everything is fine unless I saw the answer where they give the counterexample of this theorem. Now even seeing the counterexample, when I read this proof I'm not able to find any errors in it ? Can somebody point me out in the right direction ?