Statement to be proved: If $A \subsetneq B$, and $B \subsetneq C$, then $A \subsetneq C$.
First method: Inequality: If $A \subsetneq B$ then there exists an $x\in B$ such that $x\notin A$. Let $z:z\in B \wedge z\notin A$
Similarly, if $x\in B\implies x\in C$, then $z\in C$, but if $z\notin A$, then there is an element in $C$ not in $A$, so $A\ne C$.
$(x\in A\implies x\in B, x\in B \implies x\in C) \implies (x\in A \implies x\in C)$. Thus at best $x\subset C$. But since $A\ne C$, then $A\subsetneq C$. Q.E.D.
Second method (I couldn't use this in the exercise as cardinality came later):
$A\subsetneq B$ means that all elements in $A$ are in $B$, but some elements in $B$ are not in $A$. So $n(A)<n(B)$. Same can be said about $B$ and $C$: $n(B)<n(C)$. Combining the two inequalities: $n(A)<n(C)$ so it follows that $A\neq C$.
The fact that all elements in $A$ are in $C$ is proven the same way as in the first method.
I think this is correct, but I'm not entirely sure as to the logical soundess. I also think there is a faster way of doing it (but I mainly want to know if it is correct).
EDIT: I use $\subsetneq$ to mean "A is a subset of B, but A is not equal to B"