Proving or Disproving statements using sets I just don't seem to get proofs or set theory so hopefully my question makes sense. 
I'm not sure when I should or shouldn't use an example to prove or disprove a statement?
One example question is, if C $\subseteq$ A and D $\subseteq$ B, then C $\cup$ D $\subseteq$ A $\cup$ B.
I want start by making set A = $\{1, 2, 3, 4, 5\}$ , B = $\{1, 2, 3, 4, 5, 6\}$ C = $\{1, 2, 3\}$ and D = $\{1,2,3,4\}$
and this would show an example proving this statement. However I think this might be wrong because it only shows one example. 
So, I tried to think of a counterexample that would show that the statement is false. However I'm not sure if I should then try to prove, If C $\subseteq$ A and D $\subseteq$ B, then C $\cup$ D $\subseteq$ A $\cup$ B false or if I should be proving, If C $\subseteq$ A and D $\subseteq$ B, then C $\cup$ D $\subsetneq$ A $\cup$ B false??
I've also tried using x $\in$ C $\subseteq$ A, then x $\in$ C and x $\in$ A, x $\in$ D 
$\subseteq$ B, then x $\in$ D and x $\in$ B
But, I didn't know what to do from there. 
 A: In this particular case, you are asked to prove a "for all" type question.  For all sets $A, B, C$, and $D$ with $C\subseteq A$ and $D\subseteq B$ prove that $C\cup D\subseteq A\cup B$.
To show the containment, take an arbitrary element from the left and show that it is in the right.  For example, let $x\in C\cup D$, then either $x\in C$ or $x\in D$.  Without loss of generality, suppose that $x\in C$.  Then $x\in A$ since $C\subseteq A$ and hence $x\in A\cup B$.
A: Your last approach is really the way to try to go--that is, proving most set properties requires some sort of "element-chasing" proof (unless you're allowed to use simple set algebra, but it does not look like you are quite there just yet). Consider the following element-chasing proof for your claim.
Claim: If $C\subseteq A$ and $D\subseteq B$, then $C\cup D\subseteq A\cup B$. 
Proof. Choose an arbitrary $x\in C\cup D$. Then, $x\in C$ or $x\in D$. 


*

*If $x\in C$, then $x\in A$ since $C\subseteq A$; also, if $x\in A$, then surely $x\in A\cup B$.

*If $x\in D$, then $x\in B$ since $D\subseteq B$; also, if $x\in B$, then surely $x\in A\cup B$.


Either way, what you have shown is that
$$
x\in C\cup D\implies x\in A\cup B.
$$
That is all that is required to show one set is a subset of another set.
