Prove that $A \setminus (B \setminus C) =(A \setminus B) \cup C \iff C \subset A$ Prove that $A \setminus (B \setminus C) =(A \setminus B) \cup C \iff C \subset A$
I have the following:
Using $X - Y = X \setminus Y = X \cap Y^c$ into:
$A \cap (B \cap C^c)^c = (A \cap B^c) \cup C$
then 
$A \cap (B^c \cup C) = (A \cup C) \cap (B^c \cup C)$
This statement is only true if $C \subset A$. Therefore, we are done.

Is this rigorous enough proof? I know usually we solve these problems by letting some element $x$ be a member of part of the statement and continue from there, but shouldn't what I wrote be sufficient? 
 A: To me your last statement "This state is only true..." is almost as big of a jump as the original statement. 
What you "need" to write down depends on your instructor (or really your homework grader). If I had assigned such a problem, I wouldn't be content with your answer (not just because you only addressed half of the double implication). I'd like to see more on the implication you addressed.
You might want to restructure your proof more like the style you indicated ("...usually we solve...").
So I would suggest that you prove "$A \setminus (B \setminus C) =(A \setminus B) \cup C \Longrightarrow C \subset A$" first. 
This means you should assume that $A \setminus (B \setminus C) =(A \setminus B) \cup C$" and then suppose  $x \in C$ and try to use the assumed statement to get that $x \in A$. [This is easy since $x \in C$ implies that $x \in \mbox{anything} \cup C$].
Next, prove "$A \setminus (B \setminus C) =(A \setminus B) \cup C \Longleftarrow C \subset A$". To do this assume $C \subset A$ and then show $x \in A - (B-C)$ implies $x \in (A-B) \cup C$ and conversely $x \in  (A-B) \cup C$ implies that $x \in A-(B-C)$. 
Part of this might go: Suppose $x\in A-(B-C)$. Then $x \in A$ and $x \not\in B-C$. Now $x \not\in B-C$ implies that either $x \not\in B$ or $x \in C$. If $x \not\in B$, then $x \in A -B$ (since $x \in A$). Thus $x \in (A-B) \cup C$. Otherwise,  if $x \in C$ then $x \in (A-B)\cup C$. Therefore, all $x\in A-(B-C)$ also belong to $(A-B)\cup C$. Thus $A-(B-C) \subset (A-B)\cup C$.  
A: This is not directly an answer to your question, but here is a proof in an alternative style, where we first use the definitions from set theory to translate to the 'logic level', so that we can use the laws of logic.$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\followsfrom}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$
Using this approach, the left hand side of the statement then becomes:
$$\calc
    A \setminus (B \setminus C) \;=\; (A \setminus B) \cup C
\op=\hint{extensionality; definitions of $\;\setminus,\cup\;$}
    \langle \forall x :: x \in A \land \lnot (x \in B \land x \not\in C)
    \;\equiv\;
    (x \in A \land x \not\in B) \lor x \in C \rangle
\op=\hint{logic: DeMorgan -- simplifying to give both sides a similar shape}
    \tag 0
    \langle \forall x :: x \in A \land (x \not\in B \lor x \in C)
    \;\equiv\;
    (x \in A \land x \not\in B) \lor x \in C \rangle
\endcalc$$
And the right hand side is just
$$\calc
    C \subseteq A
\op=\hint{definition of $\;\subseteq\;$}
    \langle \forall x :: x \in C \then x \in A \rangle
\op=\hints{logic: $\;\lnot \phi \lor \psi\;$ is a different way to write $\;\phi \then \psi\;$}
    \hint{-- usually $\;\then\;$ is not very easy to manipulate}
    \tag 1
    \langle \forall x :: x \not\in C \lor x \in A \rangle
\endcalc$$
Now, comparing the shapes of $\ref 0$ and $\ref 1$, we see that to prove these equivalent, it suffices to prove the following statement of propositional logic:
$$
\tag 2
P \land (Q \lor R) \;\equiv\; (P \land Q) \lor R \;\equiv\; \lnot R \lor P
$$
This is in fact the logic level 'equivalent' of the original statement.
Here is a proof of $\ref 2$:
$$\calc
    P \land (Q \lor R) \;\equiv\; (P \land Q) \lor R
\op=\hints{LHS: distribute $\;\land\;$ over $\;\lor\;$}
    \hint{-- to give both sides a similar shape}
    (P \land Q) \lor (P \land R) \;\equiv\; (P \land Q) \lor R
\op=\hint{$\;\lor\;$ distributes over $\;\equiv\;$}
    (P \land Q) \lor (P \land R \;\equiv\; R)
\op=\hint{both $\;\lnot \phi \lor \psi\;$ and $\;\phi \equiv \phi \land \psi\;$ are alternatives for $\;\phi \then \psi\;$}
    (P \land Q) \lor \lnot R \lor P
\op=\hint{use negation of $\;P\;$ on other side of $\;\lor\;$}
    (\false \land Q) \lor \lnot R \lor P
\op=\hint{simplify}
    \lnot R \lor P
\endcalc$$
That completes the proof.
