$A \subset S$ then $A\cap S =A$

How would I prove the following $A \subset S$ then $A\cap S =A$

Not sure I been thinking of an example say $A={1,2,3}$ and $S={1,2,3,4,5,6}$

So I tried the following $(x:x\in A \wedge x\in S)$

so I let $x:x\in A$ I am not sure if this finishes the proof. Because I can can also $x\in S$ I think it would also be true if $(x:x\in S)$

We can prove the equality of two sets, $X=Y$ by showing $Y-X=\Phi$ and $X-Y=\Phi$. Let's try the same here.

(1) Prove that $A-(A\cap S)=\Phi$

$A-(A\cap S)$

$=\{x:x\in A-(A\cap S)\}$

$\Leftrightarrow \{x:x\in A \wedge x \notin A \cap S\}$

$\Leftrightarrow \{x:x\in A \wedge (x\notin A \lor x\notin S)\}$

$\Leftrightarrow \{x:(x\in A \wedge x\notin A) \lor (x\in A \wedge x\notin S)\}$, by the distributivity of $\wedge$ over $\lor$

$\Leftrightarrow \{x:(x\in A \wedge x\notin A) \lor (x\in S \wedge x\notin S)\}$, since by $A \subset S$, $x \in A \implies x \in S$

$=\Phi$

(2) Prove that $(A\cap S)-A=\Phi$, this is trivial (a tautology for any sets $A$, $S$)

$(A \cap S) - A$

$=\{x:x\in (A \cap S) - A\}$

$\Leftrightarrow \{x:x\in A \cap S\wedge x\notin A\}$

$\Leftrightarrow \{x:(x\in A \wedge x \in S) \wedge x\notin A\}$

$\Leftrightarrow \{x:(x\in A \wedge x\notin A) \wedge x \in S\}$, by commutativity and associativity of $\wedge$

$=\Phi$

Combining (1) and (2), we have $A=A\cap S$

I would say it's pretty easy to show this through a Venn diagram, if it helps you.

Also, if you have a set S for which $A \subset S$, then any $a\in A$ is also in S, but not every $s\in S$ is in A. So, you have:

$$A\cap S = \{s\in S | s \in A\} = \{s \in A\} = A $$

I think that's enough if you just want to understand why that is, but for more detail look here: https://proofwiki.org/wiki/Intersection_with_Subset_is_Subset

To prove that two sets are equal, you prove that they have the same elements. The elements of $A$ are, well, the elements of $A$. The elements of $A \cap S$ are, by definition, the elements of $A$ which are also elements of $S$. But all of them are elements of $S$, since that is precisely what $A \subseteq S$ means.