Prove that for all sets A,B,C if B $\subseteq$ C then A $\cup$ B $\subseteq$ A $\cup$ C I don't really know what to do with unions, can someone help me?
I assume $B\subseteq C$, (I want to show that $A\cup B \subseteq A\cup C$)
suppose $A\cup B \subseteq A\cup C$ and let $x\in A\cup C$ so that
$x\in A$ or $x\in C$...
I don't really know what to do up to this point, can someone offer me guidance?
 A: You just want to prove that every element in $A\cup B$ must be in $A\cup C$. To this end, suppose $x\in A\cup B$. By the definition of union, this means that $x\in A$ or $x\in B$. Let's consider these cases separately:
If $x\in A$, then $x\in A\cup C$.
If $x\in B$, then from $B\subseteq C$ we obtain $x\in C$ and consequently $x\in A\cup C$.
Therefore every element in $A\cup B$ is in $A\cup C$, so we have $A\cup B\subseteq A\cup C$.
A: In his comment, OP Ryan says

Thanks for the reply, I have a poor understanding of unions but your
  response definitely helps!

To promote understanding and intuition, here is a more wordy/elaborate description of what is happening. We look at
$\tag 1 B\subseteq C \Rightarrow A\cup B \subseteq A\cup C$
as a way of modifying both $B$ and $C$. In a rigorous treatment you can't say things like 'a new $B$'.
We start with
$\tag 2 B \subseteq C$
and want to find a relationship between $A\cup B$ and $A\cup C$. But the union operation is how we 'add stuff into a set'. So, anything that gets added to both the left hand side $B$ and right hand side $C$ of (2) would not change the subset relationship relationship, since it would be 'added' to both sides.
In what follows we breakdown how each element of $A$ gets added to $B$ and $C$.
If $A$ is a subset of $B$, nothing changes using the union operation,
$\tag 3 A \subseteq B \,\land\,  B\subseteq C \;\Rightarrow  A\cup B = B \,\land\, A\cup C = C$
To show that (3) is true you need to show that $A$ is also a subset of $C$.
So how do you add 'new stuff'? One way is if $B$ is a proper subset of $C$, we can take an element $c$ that is in $C$ but not in $B$, and then
$\tag 4 \{c\} \cup B \subseteq \{c\} \cup C = C$
So when we add this kind of new stuff, nothing happens to $C$ but the 'new' $B$ 'looks more like' $C$.
The last thing to consider is adding new stuff that does not appear in $C$ (and then it also can't be in $B$). So if $k$ is a completely new object, we can throw that in also, and
$\tag 5 \{k\} \cup B \subseteq \{k\} \cup C$
In this case we changed both $B$ and $C$ by adding in a new element.
In summary, if you look at (1) in a dynamic way, then as you "pour" $A$ in, 
(a) $B$ and $C$ don't change 
(b) $B$ grows larger but remains a subset of $C$
(c) Both $B$ and $C$ grow larger in the same way
(d) A mixture of (a), (b), and (c)
A: Here is a more algebraic kind of proof:
In general you have that $X \subseteq Y$ iff $X \cap Y = X$ (take a minute to see why this is true)
Applied to your case: 
We are given $B \subseteq C$, and so $B \cap C =B$
Therefore, $(A \cup B) \cap (A \cup C) = A \cup (B \cap C) = A \cup B$
So, since $(A \cup B) \cap (A \cup C) = A \cup B$, we have $A \cup B \subseteq A \cup C$
