discrete math use an element argument Q)Let U be a universe.Use an element arguement to prove the following statement.
For all sets A,B and B in P(U),(C-A) u (B-A)⊆ ( B U C) -A. 
Def : Z ⊆ W ={(z,w):x∈ X and y ∈ Y}.
Proof:
W=(C-A) U (B-A) ={(c,a):a∈A and c∈C}U{(a,b):a∈A and b∈B}
Z=(B U C)-A = {(a,y):a∈A and y∈BU C}
For all  a∈A:
Case 1.b∈C. If b∈C then (a,b)∈Z. Also (a,b)∈W.
Case 2. b∉C.If b∉C, then b is not in BU C. Then (a,b) is not in Z. b is also not in C-A, so it's not in W.
Case 3. By definition of the Union, a is also a member of C-A and A - C. By definition of intersection, a belongs to(B-A) ∩ (C-A). 
Case 4. By definition of a subset, (B-A) ∩ (C-A) is a subset of (B U C) - A. 
Case 5. Therefore (B-A) ∩ (C-A)⊆(B U C) - A.
My first time doing element argument.I spend hours to solve this, i want to make sure i am right. Can pls check and advice if i am right or wrong. And where i can make improvemet? Thanks for ur time
 A: Before anything else, here are some definitions.

$A\subseteq B$ if and only if for every $a\in A$ you have also that $a\in B$
That is to say, there is nothing that is in $A$ while not in $B$. (note, $\emptyset$ is always a subset of every other set)

Another definition which seems to have confused you

$A\backslash B = \{x~:~x\in A\text{ and }x\notin B\}$
(equivalently written as $A-B$ or $A\setminus B$ or $A\cap B^c$)

Here is the question as I understand it (given your typing).
Let $U$ be a universe.  Use an element argument to prove the following statement: For all sets $A,B$ and $C$ in $\mathcal{P}(U)$, you have that $(C\backslash A)\cup (B\backslash A)\subseteq (B\cup C)\backslash A$.
Here is an elegant set theoretic proof that uses DeMorgan's laws and symbolic manipulation, (note here that each $=$ can be interpreted left to right as $\subseteq$ while also can be interpreted right to left as $\supseteq$)
$(C\backslash A)\cup (B\backslash A) = (C\cap A^c)\cup (B\cap A^c) = (C\cup B)\cap A^c = (B\cup C)\backslash A$

Using an element argument, it should be set up as follows:
$x\in (C\backslash A)\cup (B\backslash A)\Rightarrow x\in (C\backslash A)$ or $x\in (B\backslash A)\Rightarrow (x\in C$ but not $A)$ or $(x\in B$ but not $A)\Rightarrow x\notin A\text{ and }x\in C\cup B$
$\Rightarrow x\in (B\cup C)\cap A^c\Rightarrow x\in (B\cup C)\backslash A\Rightarrow (C\backslash A)\cup (B\backslash A)\subseteq (B\cup C)\backslash A$
If you wish to do a case argument (which is much longer), then consider all possibilities for $x\in^? A, x\in^? B, x\in^? C$ (there are 8 possible configurations) and look to see if each of the eight configurations leads to $x$ being an element of the set on the left and being an element of the set on the right or $x$ not being an element on the left (with no concern about the right).
A: Here is a slightly different way to approach element arguments (a.k.a. element chasing): start at the most complex side, expand the definitions, simplify using the laws of logic, and take it from there.
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$In this case, let's calculate which $\;x\;$ are in $\;(C - A) \cup (B - A)\;$.

For all $\;x\;$,
$$\calc
x \in (C - A) \cup (B - A)
\calcop={definition of $\;\cup\;$}
x \in C - A \;\lor\; x \in B - A
\calcop={definition of $\;-\;$, twice}
(x \in C \land x \not\in A) \;\lor\; (x \in B \land x \not\in A)
\calcop{\tag{*}=}{logic: extract common conjunct $\;x \not\in A\;$}
(x \in C \;\lor\; x \in B) \;\land\; x \not\in A
\calcop={definition of $\;\cup\;$}
(x \in C \cup B) \;\land\; x \not\in A
\calcop={definition of $\;-\;$}
x \in (C \cup B) - A
\endcalc$$
By set extensionality, we've proven $\;(C - A) \cup (B - A) \;=\; (C \cup B) - A\;$.

In the above proof, the simplification is done in step $\ref{*}$: that is the key step of this proof.
Note that we've proven equality ($\;=\;$) instead of the weaker $\;\subseteq\;$, since there was no reason to use implication ($\;\Rightarrow\;$) anywhere.
Also, compare this proof to the "elegant set theoretic proof" from JMoravitz's answer: it has exactly the same structure, but that proof is on the set theory level, and this one is on the element level.
