Why does proving a set is a subset of another only require you to take a single element of the subset? I don't really understand why to show $A\subseteq B$, all we need to show is that $\exists x \in A$ and $x \in B$? 
Doesn't this notation mean that there is a specific element of A inside B, not arbitrary element of A is within B? Hence, you should say $\forall x \in A$, $x \in B$?
 A: It is not true.
By your way if $A\subseteq B$ then $B\subseteq A$, because $\exists x\in A;x\in B\implies \exists x\in B;x\in A$. Further more, let's take a simple counter example, $A=\{1,2\},B=\{1,3\}$ the statement $\exists x\in A;x\in B$ is true,  exists a member of the set $A$ that is also in the set $B$, but $A$ is not a subset of $B$.
The definition of subset is $\forall x(x\in A\implies x\in B)$ so for every arbitrary $x$ you will take if it is in $A$ it will be also in $B$ so the way to prove subset is $\forall x\in A;x\in B$.

Sometimes people say $x\in A\implies x\in B$, this way also works because here $x$ is arbitrary, so it is equivalent to the definition
A: I'm not sure where you've seen that $\exists x \in A (x \in B)$ is equivalent to $A \subseteq B$, but that's false. For $A$ to be a subset of $B$ means that for every element $x$ of $A$, the element $x$ is also an element of $B$, i.e. $\forall x (x \in A \to x \in B)$. 
In practice, we show that $A \subseteq B$ by taking a generic element of $A$ and showing it's also an element of $B$. This is possible because the sets $A$ and $B$ are usually given by certain defining formulas. For example, $A$ might be the set $\{m \in \mathbb{N} | \exists n \in \mathbb{N}(m=2n \wedge n \geq 10) \}$, i.e. the set of even numbers greater than or equal to $20$. If we want to show $A \subseteq \{20, 21, 22, ...\}$ we simply take an arbitrary $x \in A$ and show it satisfies the defining condition of $\{20,21,22,...\}$.
A: The single element is an arbitrary element of the set you are proving is a subset...
You do not prove $\exists x\in A : x\in B$...
Rather, you prove that $\forall x\in A$, it is true that $x\in B$... 
