Let's look at how the subset is defined.
$A \subseteq B$ means that $\forall x: x \in A$ $\Rightarrow x \in B$.
Let's prove this statements by a chain of If and only if statements.
- $A \subseteq B$ $\Leftrightarrow$ $\forall x: x \in A \Rightarrow x \in B$
Now, we can write $A \Rightarrow B$ using only AND-OR-NOT formula like below.
$A \Rightarrow B$ IFF $\neg A \vee B$. (You can simply check this using truth table).
- $\forall x: x \in A \Rightarrow x \in B \:$ IFF $\: \forall x: \neg(x \in A) \vee (x \in B) \:$ IFF $\:\forall x: \neg((x \in A) \wedge \neg(x \in B))\:$ IFF $\:\forall x: \neg((x \in A) \wedge (x \notin B))\:$ IFF $\:\forall x: \neg((x \in A) \wedge (x \in B^c))\:$ IFF $\:\forall x: \neg(x \in A \cap B^c))\:$ IFF $\:\forall x: (x \notin A \cap B^c))$.
This is the end. We've showed that if $A \subseteq B$, then $\forall x: (x \notin A \cap B^c))$, which means that there is not any element in the intersection of $A$ and $B^c$, which is in effect in empty.
$\blacksquare$
Your proof is correct. Here is a proof that avoids any mention of specific elements (following the theme of my answer to one of your previous questions). The key statements we use are the following:
(a) If $X$ and $Y$ are sets then $X \subseteq Y$ iff $X \setminus Y = \emptyset$.
(b) If $X$ and $Y$ are sets then $X \cup Y = \emptyset$ iff $X = \emptyset$ and $Y = \emptyset$.
(We discussed both of these before, so let's not reprove them!)
Now, in this problem we care about when $C \subseteq A \Delta B$. So, guided by property (a), we should examine $C \setminus (A\Delta B)$. Use axioms of set operations (e.g., De Morgan etc) to prove:
$$
C \setminus (A\Delta B) = \big(C \setminus (A\cup B)\big) \cup \big(A \cap B \cap C\big)\tag{1}
$$
I have hidden the proof of $(1)$ at the bottom of this answer; but try it yourself first. It's also a sensible thing to say out loud: $A \Delta B$ is the set of elements that are in either $A$ or $B$, but not both. So being in $C \setminus (A \Delta B)$ is the same as either being in $C$ and not in $A$ or $B$, or being in $C$ and in both $A$ and $B$.
Once you have $(1)$, the rest is very straightforward.
\begin{align}
C \subseteq A \Delta B &\iff C \setminus (A \Delta B) = \emptyset \tag{using (a)} \\
&\iff \big(C \setminus (A\cup B)\big) \cup \big(A \cap B \cap C\big) = \emptyset \tag{using (1)}\\
&\iff C \setminus (A \cup B) = \emptyset \text{ and } A \cap B \cap C = \emptyset \tag{using (b)}\\
&\iff C \subseteq A\cup B \text{ and } A\cap B\cap C = \emptyset \tag{using (a)}
\end{align}
Proof of $(1)$:
Recall that $$A \Delta B = (A \cup B) \setminus (A \cap B) = (A \cup B) \cap \neg(A \cap B)\tag{2}$$
So \begin{align}C \setminus (A \Delta B) &= C\cap \neg\big((A \cup B)\cap \neg (A \cap B)\big) \tag{by (2)} \\ &= C \cap \big(\neg (A \cup B) \cup (A \cap B)\big) \tag{De Morgan} \\ &= \big(C \cap \neg (A \cup B)\big) \cup \big(C \cap (A \cap B)\big) \tag{distributivity} \\ &= \big(C \setminus (A \cup B)\big) \cup \big(A \cap B \cap C\big)\end{align}
In the last line we used the definition of set difference on the left side, and associativity/commutativity of intersection on the right side.
Best Answer
Try mutual subset inclusion.
($\to$): Suppose $x\in A\subseteq B$. Then $x\in A\to x\in B$; that is, $x\not\in A\lor x\in B$. The negation of this is $x\in A\land x\not\in B$; that is, $x\not\in A\cap \overline{B}$.
The other direction is trivial since you are dealing with the empty set.