Show that if $A$, $B$, and $C$ are sets such that $|A| \leq |B|$ and $|B| \leq |C|$, then $|A| \leq |C|$. If $|A|\leq|B|$, then $A \subset B$ and there exists $x \in B$, $x \not\in A$. 
If $|B|\leq|C|$, then $B \subset C$ and there exists $x \in C$, $x \not\in B$.
Therefore there exists $x \in C$, $x \not\in A$. So, $A \subset C$ and $|A|\leq|C|$.  
Would this be a correct presentation to satisfy the question?
 A: So first, let's say that $|A| \leq |B|$ iff there is an injection $f : A \to B$.  (So it could be the case that $A \subseteq B$ but in general that is not true).  Now, we can prove the proposition quite simply.
Since $|A| \leq |B|$ that means there is a injection $f: A \to B$ and since $|B| \leq |C|$ that means there is an injection $g : B \to C$.  So now, to prove that $|A| \leq |C|$ what do we need?  We need an injection $h: A \to C$.  But what if we try composing the functions in this order $g \circ f$ (to remember what to do first read $\circ$ as "after", so we first do $f$ then $g$).  This will give us a function with domain the same as $f$ (notice the domain of $f$ is $A$) and with codomain the same as $g$ (notice the codomain of $g$ is C).  So let $h = g \circ f$, and then $h :A \to C$.  Now all that is left to prove is that the composition of injections is indeed an injection, but this is a good exercise so I will let you try it!  If you get stuck see here (but try it first!)
A: No, if $A=\{1\}$ and $B=\{2,3\}$. We have $|A|\le |B|$ but it no means that $A \subset B$.
