proof-explanation : if $A \subseteq B$ and $B \subseteq C$ , Prove that $A\subseteq C$ I have looked quite a few examples regard to the proof of above. they are all similar : 
Let  $x\in A$  
since $A \subseteq B$ , $x\in A \implies x \in B$
since $B \subseteq C$ , $x\in B \implies x \in C$
therefore   $A\subseteq  C$    //This is where I don't get it
how could they  come to the conclusion that $A ⊆ C$  based on $x ∈ C$ ? I mean $x$ is only an element. I can easily give a counter example to disprove it. Let's say $x = 1$, $A = \{ 1,2 \}, B = \{ 1, 2, 3\},  C = \{1,3,7\}$, In this case , $x ∈ A, x ∈ B, x ∈ C$, but It doesn't show that $A ⊆ C$.  However , you might notice the example that $B$ is not a subset of $C$ which contradict the given condition. If that's the case, the question itself is already a proof. why do we need to prove that $x ∈ B$ and $x ∈ C$?
I think I'm missing something, But I don't know what it is , Would you please provide details as much as possible.
 A: Suppose $A\subseteq B$ and $B\subseteq C$.
If $A\subseteq B\Longrightarrow\forall x\in A,~x\in B$.
If $B\subseteq C\Longrightarrow\forall x\in B,~x\in C$.  
Directed:
So by transitivity $\forall x\in A\Longrightarrow x\in B\Longrightarrow x\in C\Longrightarrow A\subseteq C$.
Absurd:
Suppose $A\not\subseteq C\Longrightarrow\exists x\in A:x\notin C$.
But if $x\in A\Longrightarrow x\in B$, but if $x\in B\Longrightarrow x\in C$.
So $x\in C~\wedge x\notin C$, then the hypothesis was false, so $A\subseteq C$.
A: The point is that not only for a single element, being in $A$ implies being in $C$, but as the arguument is valid for any $x$, we can have inclusion.
A: This shows that no matter which element of $A$ you choose, it is also an element of $C$. So each element of $A$ is an element of $C$ and hence $A\subseteq C$
A: Indeed your counter example won't disprove the statement in the question title, as the $B\subseteq C$ assumption is key. 
You have that, for any $x\in A$, we have $x\in B$ since $A\subseteq B$. You also have that, for any $y\in B$, we have $y\in C$ since $B\subseteq C$. So, any element $x\in A$ is also in $C$, since every element of $A$ is in $B$, and every element of $B$ is in $C$. Showing that, for an arbitrary $x\in A$ we have $x\in C$ is precisely what it means for $A$ to be contained in $C$.
A: Suppose $A $ is not a subset of $C$.
this means there exist $a \in A $ which is not in $C $.
$A\subset B $ and $a \in A $
$$\implies a\in B. $$
$a\in B $ and $B\subset C $
$$\implies a\in C $$
and this is the contradiction.
