inclusion of sets -transitive? show that if $A\subseteq B$ and $B\subseteq C$ then $A\subseteq C$
Can I do it with using injective functions?
$A\subseteq B$  means there exists an injective fcn $f:A\to B$
$B\subseteq C$  means there exists an injective fcn $g:B\to C$
then the composition  $g\circ f:A\to C$ is also an injective function then $A\subseteq C$
in each case all the functions are identity functions
 A: Caution: The existence of an injection between $A$ and $C$ doesn't necessarily imply $A \subseteq C$. 
For example, consider set $A$, the set of all even integers, and set $B$, the set of all odd integers. Certainly, there exists an injection: $f: A \to C$, $\;f(x) = x + 1\;$ (which is not only injective, but surjective, as well). But clearly, $\;A \nsubseteq C$.
The converse is true: if $A \subseteq C$, then an injection $h: A \to C$ exists. 

But we can easily prove the inclusion $A \subseteq C$ by "element chasing:"  a standard way to prove set inclusions, and/or  set equivalencies.
We have $A \subseteq B$ and $ B \subseteq C$. And we want to prove that this necessarily implies $A\subseteq C$. 


*

*$(1)$ Suppose $x \in A\quad $ (Assumption) 

*$(2)$ We know $A \subseteq B$ means $x \in A \implies x \in B.\;$ So given $(1)$, we have $x \in B$. 

*$(3)$ We know $B \subseteq C$ means $x \in B \implies x \in C$. So given $(2)$, we have $x \in C$.


$(4)\;\;x \in A \implies x \in C$. $\quad[(1) - (3)]$
Therefore, $A \subseteq C$.
A: No, you can't do so this way.  
"A⊆B means there exists an injective function f:A→B" is false.  
"A⊆B implies there exists an injective function f:A→B" is true.  
"A⊆B means there exists an injective function f:A→B" would only come as true if both
(1) "A⊆B implies there exists an injective function f:A→B" and 
(2) "the existence of an injective function f:A→B implies A⊆B" were true also.  But, (2) is false, and thus "A⊆B means there exists an injective function f:A→B" is false. 
A: Well, you say
if x ∈ A, then x ∈ B because of A ⊆ B,

but a distinction between x as a "hard element" and X as a set itself is not substancial here. And if you write your conclusion as 
if X ⊆ A, then X ⊆ B because of A ⊆ B

than, you'll see that your reasoning, is exactly, what you would like to prove.
