Is the proof of $(A\cap B) \subseteq A$ correct? I need to proof that: $(A\cap B) \subseteq A$ is true.
My attempt was:
$$(A\cap B) \subseteq A = \{\forall x: (x\in (A\cap B)) \to (x \in A)\}$$
$$= \{\forall x: ((x\in A) \land (x\in B)) \to (x\in A)\}$$
Using the tautology $p \land q \to p$
the statement is always true.
Is this correct or there are some observations to improve?
 A: It is a somewhat convincing proof by equivalent substitutions.  
However you should not be using set builder notation.  
Moreover you are working backwards: Instead you need to demonstrate that an axiomatic tautology entails that which is to be proven.
$$\begin{align}&p\wedge q\to p&&\text{is a tautology}\\&x\in A\wedge x\in B\to x\in A&&\text{for arbitrary $x$, by substitution}\\&x\in A\cap B\to x\in A&&\text{for arbitrary $x$, by definition of intersection}\\&\forall x~(x\in A\cap B\to x\in A)&&\text{by generalisation.}\\ &A\cap B\subseteq A&&\text{by definition of subset.}\end{align}$$
Therefore $A\cap B\subseteq A$ is a theorem.
$\blacksquare$
A: $\bullet\space$ Inclusion is defined via a universally quantified conditional 

set $X$ is included in set $Y$ 
iff the following holds: 
$\forall x ( x\in X \rightarrow x\in Y)$

$\bullet\space $ Herer we want to prove that this sentence is true with : $X = A\cup B$ and $Y = A$ 
$\bullet\space $ Consider an arbitrary object $a$. 
$\bullet\space$ Suppose, in view of conditional proof, that $a\in X$ is true, that is, here, that 
$$a\in A\cap B$$ is true. 
Note : we are not asserting that $a\in A\cap B$ is true; we are simply putting ourselves in a  possible situation in which this condition holds, in order to see what would follow in that case . 
$\bullet\space$ By the definition of intersection ($\cap$), the above hypothesis means that ($a\in A \land  a\in B$) is true, which implies, by "AND-Elimination" rule , that 
$$a\in A$$
is true. ( In other words, under our hypothesis that $a\in X$ is true, we have reached our goal : $a\in Y$). 
$\bullet\space$ The Condiitonal proof inference rule allows us  to derive : 
$$a\in A\cap B \rightarrow a\in A $$
$\bullet\space$ But, object $a$ was arbitrary, meaning that everything that has been deduced regarding $a$ also holds for any object. We may therefore use Universal Generalization to infer : 
$$\forall x ( x\in A\cap B \rightarrow x\in A)$$. 
$\bullet\space$ It means that the universally quantified conditional 
$$\forall x ( x\in X \rightarrow x\in Y)$$
holds with $X = A\cap B$ and $Y=A$; and therefore , by the definition of inclusion, that 
$$A\cup B \subseteq A$$
Note : 
"AND-Elimination" is the rule that says " From a conjunction , any conjunct may be inferred". 
Conditional Proof is a propositional logic inference rule saying : " if you can show that proposiiton $Q$ holds under the hypothesis that $P$ is true, then, you have actually proved that the conditional $P\rightarrow Q$ is true" 
