Prove that if A is a subset of B intersect C, then A is a subset of B and C. Show why this doesn't work for union. I try to prove that if $A \subset B \cap C$, then $A \subset B$ and $A \subset C$.
I go through:
(1) For all $x \in A, x \in B \cap C$.
(2) For all $x \in A, x \in B$ and $x \in C$
(3) For all $x \in A, x \in B$ and for all $x \in A, x \in C$
(4) So $A \subset B$ and $A \subset C$
I'm not sure if step 2 to 3 is sound. But my real question is why I can't replace all "$\cap$"s with "$\cup$"s and all "and"s with "or"s and prove the obviously false $A \subset B \cup C \implies A \subset B$ or $A \subset C$.
 A: Suppose $A \subseteq B \cap C$, and take an arbitrary element $x \in A$. Since $A \subseteq B \cap C$, you know that $x \in B \cap C$, so that $x \in B$ and $x \in C$. As such, from the hypothesis that $A \subseteq B \cap C$, from the sole knowledge that $x \in A$, you can deduce that $x \in B$ (and hence $A \subseteq B$) and $x \in C$ (and hence $A \subseteq C$).
However, suppose now that $A \subseteq B \cup C$. If you wanted to prove that $A \subseteq B$ or $A \subseteq C$, you'd need every element of $A$ to be an element of $B$, or you'd need every element of $A$ to be an element of $C$. But given the hypothesis, you know that for each $x \in A$, either $x \in B$ or $x \in C$, but you don't know which—it's possible for some elements of $A$ to be in $B$, and some elements of $A$ to be in $C$, without necessitating that each element of $A$ be in only one of the sets $B$ or $C$. So the argument doesn't carry over.
And, indeed, there are counterexamples; take $A = \{ 0, 1 \}$, $B = \{ 0 \}$ and $C = \{ 1 \}$. Then $A \subseteq B \cup C$, but $A \nsubseteq B$ and $A \nsubseteq C$.
The crux of why you can't just replace $\cap$ by $\cup$ has to do with a duality in propositional logic: if $p \to (q \wedge r)$, then $p \to q$ and $p \to r$; but $p \to (q \vee r)$ does not imply $p \to q$ and it does not imply $p \to r$.
A: Yes.   You can also express it thusly:
$$\begin{array}{l:l}(0) & A\subseteq (B\cap C) \\ (1) &\forall x~(x \in A \to x \in B \cap C)\\(2) &\forall x~(x \in A \to (x \in B\wedge x \in C))\\(2.5) &\forall x~((x \in A \to x \in B)\wedge (x \in A\to x \in C))\\(3) &\forall x~(x \in A \to x \in B)~\wedge~\forall x~(x \in A\to x \in C)\\(4) & \text{So $A \subseteq B$ and $A \subseteq C$}\end{array}$$
For union, we start the same:
$$\begin{array}{l:l}(1) &\forall x~(x \in A \to x \in B \cup C)\\(2) &\forall x~(x \in A \to (x \in B\vee x \in C))\\(2.5) &\forall x~((x \in A \to x \in B)\vee (x \in A\to x \in C))\end{array}$$
But we cannot go any further because $\forall x~(P(x)\vee Q(x))$ does not infer that $(\forall x~P(x))\vee( \forall x~Q(x))$.
("All royal women are princesses or queens" does not infer that "All royal women are princesses, or all royal women are queens".)
A: Your argument is sound.  What fails in the case of union is the passage from $1$ to $2$.  In $2$ you would get $x \in B$ or $x \in C$.  You could have some of $A$ in $B$ and the rest of $A$ in $C$, so you can't reach step $4$.  An example is $A=\{1,2\}$, $B$ is the even numbers and $C$ is the odd numbers.
A: If you would try the same proof for unions, you get the following:
Choose $x\in A$. Since $A\subset B\cup C$ we have $x\in B\cup C$. Hence $x\in B$ or $x\in C$. Hence we get the statement:
$$\forall x\in A: (x\in B \text{ or } x\in C).$$
This is not equivalent to $$\forall x\in A: x\in B \text{ or } \forall x\in A:x\in C.$$
Thus this step doesn't work anymore.
