Why does $A\subset B$ and $A\subset C$ imply $A\subset B\cup C$? Suppose that $A\subset B$ and $A\subset C$. Why does this imply $A\subset B\cup C?$
If $x\in A$, then since $A\subset B$ and $A\subset C$, we know $x\in B$ and $x\in C.$ This implies $x\in B \cap C.$
But it is not generally true that $B\cap C\subset B\cup C $ since $B$ and $C$ may be disjoint.
So why does $A\subset B$ and $A\subset C$ imply $A\subset B\cup C?$
 A: The following result is useful to have at hand:
Proposition
Given some set $U$ as well as $X\subseteq U$ and $Y\subseteq U$, $X\subseteq Y$ iff $X\cap Y = X$.
Solution
Since $A\subseteq B$ and $A\subseteq C$, we conclude that $A\cap B = A$ and $A\cap C = A$.
Having said that, according to the properties of the operations on sets, we get that:
\begin{align*}
A\cap(B\cup C) & = (A\cap B)\cup(A\cap C)\\\\
& = A\cup A\\\\
& = A
\end{align*}
\begin{align*}
\end{align*}
Hence we conclude that $A\subseteq B\cup C$, and we are done.
REMARK
You can also apply the transitivity property. Indeed, we have
\begin{align*}
A\subseteq B \subseteq B\cup C \Rightarrow A\subseteq B\cup C
\end{align*}
Another related result (perhaps a stronger one) says that $A\subseteq B\cap C$.
Indeed, we can proceed as follows:
\begin{align*}
A\cap(B\cap C) & = (A\cap B)\cap C\\\\
& = A\cap C\\\\
& = A
\end{align*}
Hopefully this helps!
A: This is trivial if $A=\varnothing$.
Let $x\in A$. If $A\subset B$, then $x\in A$ implies $x\in B$. If $x\in B$, then certainly either $x\in B$ or $x\in C$. Thus $x\in B\cup C$. Thus $A\subset B\cup C$.
A: Continuing from where you left off:
\begin{aligned}x&\in B\cap C\\&\subset B\cup C.\end{aligned}
Therefore, since $x\in A\implies x\in B\cup C,$ $$A\subset B\cup C.$$
