Proof of $A \subseteq B \Leftrightarrow A \cap B = A$ (Check chain of implications) Prove $A \subseteq B \Leftrightarrow A \cap B = A$.
My attempt:
Case $\Rightarrow$:
$$\begin{align}
A \subseteq B & \Rightarrow & [x\in A \Rightarrow x\in B]  \\
&\Rightarrow &[x \in A \Rightarrow x\in A \text{ and } x\in B] \tag{1} \\
&\Rightarrow &[x\in A \Rightarrow x\in A \cap B]  \\
&\Rightarrow & A \subseteq A\cap B 
\end{align}$$
$$\begin{align}
A \subseteq B & \Rightarrow & [x\in A \Rightarrow x\in B]  \\
& \Rightarrow & [(x\in A \text{ and } x\in B) \Rightarrow x \in A] \tag{2} \\
& \Rightarrow & [x \in A \cap B \Rightarrow x \in A] \\
& \Rightarrow & A \cap B \subseteq A 
\end{align}$$
$$\begin{align}\therefore  A \subseteq B & \Rightarrow & A \subseteq A \cap B \text{ and } A \cap B \subseteq A \\
&\Rightarrow& A \cap B = A
\end{align}$$
Case $\Leftarrow$:
$\begin{align}
A \cap B = A & \Rightarrow & A \cap B \subseteq A \text{ and } A \subseteq A \cap B \\
& \Rightarrow & [(x \in A \text{ and } x \in B) \Rightarrow x \in A ] \text{ and } [x\in A \Rightarrow & (x \in A \text{ and } x \in B)] \\
& \Rightarrow & [x\in A \Rightarrow x \in B] \tag{3}\\
& \Rightarrow & A \subseteq B
\end{align}$
Is my proof correct? I am particularly not sure about line 1,2 and 3 since I just made those up (while making sure that the chain of implications is still true) as I already know the conclusion I want to arrive at.
 A: $A \cap B \subseteq A$ is always true, so it's a bit misleading to say that it is implied by $A \subseteq B$ (although not incorrect).  You can simply start with the statement:
$$x \in A \text{ and } x \in B \implies x \in A$$
to prove it$-$this is probably what you meant by the statement $(2)$.
For statement $(3)$ it might be more clear how you arrived at this if you say that $A \cap B = A$ implies $A \subseteq A \cap B$, and since $A \cap B \subseteq B$, this means $A \subseteq B$.
The rest of the proof is correct.
A: I think you waved past a couple of steps in "Case $\Leftarrow$":
$$\begin{align}
A \cap B = A &\Rightarrow A \cap B \subseteq A \land A \subseteq A \cap B \tag{1}\\
& \Rightarrow [(x \in A \text{ and } x \in B) \Rightarrow x \in A ]\land[x\in A \Rightarrow  (x \in A \land x \in B)]\tag{2} \\
& \Rightarrow  [x\in A \Rightarrow x \in B] \tag{3}\\
& \Rightarrow  A \subseteq B\tag{4}
\end{align}$$
After unpacking $A \cap B = A$ to get $(1)$ and $(2)$ (which are both perfectly correct), I think you need more work (or more justification) before concluding: $$x\in A \rightarrow x\in B\tag{3}$$ 

$$\begin{align}
A \cap B = A &\Rightarrow A \cap B \subseteq A \land A \subseteq A \cap B \tag{1}\\
& \Rightarrow [(x \in A \land x \in B) \Rightarrow x \in A ]\\ &\quad \land x\in A \Rightarrow  (x \in A \land x \in B)]\tag{2} \\
&\Rightarrow x\in A \Rightarrow (x \in A \land x\in B)\tag{3}\\ \\
& \quad \text{Assumption:} x\in A\tag{4}\\
&\qquad\quad {\small \Rightarrow} x\in A \land x \in B \tag{(5): 3 & 4}\\
&\qquad\quad {\small \Rightarrow} x\in B\tag{(6): from 5}\\\\
&\Rightarrow x\in A \Rightarrow x\in B\tag{(7): 4-6}\\ 
& \Rightarrow  A \subseteq B\tag{8}
\end{align}$$
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$For comparison, here is a different (more "logical" :-) approach.
Assuming that you're only allowed to use the definitions of $\;\subseteq\;$ and $\;\cap\;$, any proof $\;A \subseteq B \;\equiv\; A \cap B = A\;$ is essentially a proof of the equivalent
$$
\langle \forall x :: x \in A \Rightarrow x \in B \rangle
\;\equiv\;
\langle \forall x :: x \in A \land x \in B \;\equiv\; x \in A \rangle
$$
That statement directly follows from the following more general law of propositional logic: for any $\;P,Q\;$,
$$
P \Rightarrow Q \;\;\equiv\;\; P \land Q \;\equiv\; P
$$
There are numerous ways to prove this, depending on what laws of logic you are allowed to use.  For example, one can work from right to left:
$$\calc
P \land Q \;\equiv\; P
\calcop{\equiv}{split $\;\equiv\;$ into both directions -- to introduce $\;\Rightarrow\;$ as in our goal}
(P \land Q \Rightarrow P) \;\land\; (P \Rightarrow P \land Q)
\calcop{\equiv}{use $\;P \equiv \text{true}\;$ from left hand side of $\;\Rightarrow\;$ on right hand side, twice}
(P \land Q \Rightarrow \text{true}) \;\land\; (P \Rightarrow \text{true} \land Q)
\calcop{\equiv}{simplify: left part is $\;\text{true}\;$; $\;\text{true} \land R \equiv R\;$, twice}
P \Rightarrow Q
\endcalc$$
