Prove that $A\cap B\subseteq A$ I am unsure if I got it right, I proved it as follows:
$$A\cap B \longleftrightarrow x\in A \text{ and } x\in B \text{ is true }\forall x\in A \longleftrightarrow \subseteq A$$
Althought I kind of doubt the last part about it being true for all $x\in A$. Am I missing something?
I apologize in advance if this post is a duplicate.
 A: You are done after the first $\longleftrightarrow$ . This means that $x \in A \cap B \implies x \in A$, which satisfies the result by the definition of a subset.
A: Definition: $x\in A\cap B$ if and only if $x\in A$ and $x\in B$.
If it is false that $A\cap B \subseteq A$, then $\exists x\in A\cap B: x \notin A$. But this contradicts the definition. So, it must be that $A\cap B \subseteq A$.
A: By definition, $C \subseteq A$ iff $x \in C \implies x \in A$.
Apply this to $C = A \cap B$:
$x \in A \cap B \implies x \in A \text{ and } x \in B \implies x \in A$
A: Sketch of the proof. We want to prove that one set is a subset of other set. Recall the definition of a subset.

Definition. Let $A$ and $B$ be any sets. We say that $A$ is a subset of $B$, and we write $A \subseteq B,$ if every element of $A$ is, also, an element of $B,$ i.e.,
$$(\forall x) x \in A \implies x \in B$$

Hence, to prove that $A \cap B \subseteq A$ we must prove the above implication, i.e., prove that for an arbitrary objet c we have that $c \in A \cap B \implies c \in A.$

Proof. Let $x$ be an arbitrary object. Suppose that $x \in A \cap B.$ By definition of set intersection, we have that $x \in A$ and $x \in B.$ In particular, we have that $x \in A.$ Hence, by definition of subset, we conclude that $A \cap B \subseteq A.$ $\square$

Remark. Also, (and maybe to get some intuitive sense), note that if you have the propositions $p$ and $q$ and you know that $p \wedge q$ is true, what can you say about each $p$ and $q?$
They have to be both true (by definition of conjunction). Suppose that (at least) one of them is false. Without loss of generality, suppose that $p$ is the one which if false. Hence, $p \wedge q$ would be false, which is an absurd because it contradict the fact that $p \wedge q$ is true. The absurd came from assuming that (at least) one of them is false. Hence, both $p$ and $q$ must be true.
From this discussion, observe that $x \in A cap B$ is logically equivalent to $x \in A \wedge x \in B.$ Hence, if this proposition is true, then both $x \in A$ and $x \in B$ are true. Therefore, $A \cap B \subseteq A$ and $A \cap B \subseteq B.$
