Prove that $A \subseteq B$ if and only if $A \cap \overline{B}=\emptyset.$ Prove that $A \subseteq B$ if and only if $A \cap \overline{B}=\emptyset.$

Proof:
Since $A \cap \overline{B}$ implies that $x \in A$ but $x \notin B$, whilst $A \subseteq B$ implies that $x \in A$ and $x \in B$, we have a contradiction, since $x \in B$ and $x \notin B$. Thus, $A \cap \overline{B}=\emptyset.$

Is this sufficient? How else can this be proved?
 A: 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. 
A: You have to prove an "if and only if" statement.  When you have to prove "$p$ if and only if $q$", it always means you have to prove two statements.  You need to prove that $p \implies q$ and $q \implies p$ (where $p$ is the statement $A \subseteq B$ and $q$ is the statement $A \cap \overline{B} = \emptyset$).

Let's prove $p \implies q$ first.  To prove $p \implies q$, we need to assume $p$ is true, and prove $q$ is true.  So we need to assume $A \subseteq B$.  Let's prove $A \cap \overline{B} = \emptyset$.  It seems like the easiest way to prove this is by contradiction.  Suppose that $A \cap \overline{B} \neq \emptyset$.  Then there is some $x \in A \cap \overline{B}$.  That means there is some $x$ such that $x \in A$ and $x \not \in B$.  But we assumed $A \subseteq B$, which means for all $y \in A$, $y \in B$.  But we just found an element $x$ in $A$ that is not in $B$.  So we have an element that is both in $B$ and not in $B$, which is a contradiction.  Thus, $A \cap \overline{B} = \emptyset$, as desired.

Now let's prove the $q \implies p$ direction.  We have to assume $q$ and prove $p$.  Suppose $q$ is true, i.e., $A \cap \overline{B} = \emptyset$.  Let's prove $A \subseteq B$.  To prove this, we need to show if $x \in A$, then $x \in B$.  Let $x \in A$.  We know $A \cap \overline{B} = \emptyset$, which means if $x \in A$, $x$ can't be in $\overline{B}$.  But elements are either in $B$ or $\overline{B}$, since these two sets are complements of each other.  That means $x \in B$, which is what we wanted to show.  So $ A \subseteq B$, as desired.
A: First $A\subset B \Longrightarrow\overline{B}\subset \overline{A}\Longrightarrow A\cap \overline{B}\subset A\cap\overline{A}=\emptyset$ and hence $A\cap \overline{B}=\emptyset$. Conversely, $A\cap \overline{B}=\emptyset$ implies $A=A\cap(B\cup \overline{B})=(A\cap B)\cup(A\cap\overline{B})=A\cap B \Longrightarrow A\subset B$.
A: $ (\Rightarrow) $; Suppose $ A\subseteq B $. Then $ \forall x\in A,x\in B $. Therefore $ \forall x\in A,x\notin B^c $. So $ A\cap B^{c}=\varnothing $.
$ (\Leftarrow) $; Conversely suppose  $ A\cap B^{c}=\varnothing $. Therefore $ \forall x\in A,x\notin B^{c} $. Hence $ \forall x\in A,x\in B $. So we have that $ A\subseteq B $.
Here I use $B^c$ for your $\overline{B}$ :)
A: If $A\subseteq B$, then $x\in A \implies x \in B$, and if $x\in B$, then $x\not\in B^c$. So no $x$ be be in $A$ and $B^c$ at the same time, so $A\cap B^c=\emptyset$.
If $A\cap B^c=\emptyset$, we have $B\cup B^c=\omega$, where $\omega$ represents the entire space. If $x\in A, x\not\in B^c$, and so $x\in B$. Therefore $A\subseteq B$.
