$A\subset (B\cup C)$ $\implies$ $A\subset B $ or $A\subset C$ Clearly, it is not true that $A\subset (B\cup C)$ implies $A\subset B $ or $A\subset C$, since if $A=\{b,c\}, B=\{a,b\}, \text{ and }C=\{c,d\}$, then $A\subset B\cup C =\{a,b,c,d\}$ but $A\not\subset B$ and $A\not\subset C.$
However, if one tries to prove the above statement, where does the proof go wrong?
Proof:
Let $x\in A$. Since $A\subset B\cup C, x\in A\implies x\in B\cup C\implies x\in B$ or $x\in C. $ Since $x\in A\implies x\in B$ or $x\in C$ is equivalent to $x\in A\implies x\in B$ or $x\in A\implies x\in C$, this shows $A\subset B$ or $A\subset C. \blacksquare$

Note that what you have proved is $$\forall x(x\in A\implies(x\in B\lor x\in C)),$$ where as what you have to prove to show $A\subset B$ or $A\subset C$ is $$(\forall x(x\in A\implies x\in B))\lor(\forall x(x\in A\implies x\in C)).$$
The lesson is, be careful where the quantifiers ($\forall$ and $\exists$) are! Wrong placements can and do change meaning.
