Prove $A\cap A = A$. How can $A\cap A = A$ be proved?
Here is my approach-

$A \cap A$
$= \{x : x \in A \cap A \}$
$= \{x : x \in A \wedge x \in A \}$
$= \{x : x \in A \}$
$= A$

I curious to know if there exists more ways to prove the aforementioned statement apart from my approach ?
A detailed proof would be helpful.
You can just look at what intersection, i.e. $\cap$, produces for a set. Let $X,Y$ be sets, then

$$X\cap Y=\{z\mid z\in X\text{ and }z\in Y\}$$

Thus

$$A\cap A=\{x\mid x\in A\text{ and }x\in A\}=\{x\mid x\in A\}=A$$
The set $A \cap B$ is definied as $\{x: x \in A \text{ and } x \in B\}$

The set $A \cap A$ is, by consequence, $\{x:  x \in A \text{ and } x \in A\}$.

The set $A$ is obviously $\{x: x \in A\}$

Now, given any element $G \in A$, $G \in A
\cap A$, as the definition suggests.

Also, for any element $G \in A \cap A$, $G \in A$ has to be satisfied.

Thus, every element of set $A$ is also in the set $A \cap A$ and vice versa

That should be enough to declare the sets equal.