In short, the task was to prove the statement:
If $A \times B \subset B \times C$, then $A \subset C$, with the constraint $B \neq \emptyset$.
Attempted proof:
First of all, if $A = \emptyset$, then $A \times B = \emptyset$. As the empty set is a subset of any set, both parts of the implication are true.
If $A \neq \emptyset$, we can consider any $(x,y) \in A \times B$. If $A \times B \subset B \times C$, then $(x,y) \in B \times C$ as well.
From this we see that $x \in A \land x \in B$, so per the definition of subsets $A \subset B$. We can also see that $y \in B \land y \in C$, so similarly $B \subset C$.
By the transitive property of subsets, if $A \subset B$ and $B \subset C$, then $A \subset C$.
However, the above proof was dismissed, so I'm curious as to what the correct approach would be, and what faulty assumptions the proof above makes?
I think the only issue is that, since $B\ne\emptyset$, pick a $y_{0}\in B$, and start arguing that if $x\in A$, then $(x,y_{0})\in A\times B$... then you are going to have $x\in C$ at the end.
The reason for being dismissed, as I guess, is that you have written $(x,y)\in A\times B$ at the very beginning, one may ask, what if $B=\emptyset$, is that meaningful to write $(x,y)\in A\times B$ at start.
And one point further, if $A=\emptyset$, then $A=\emptyset\subseteq C$ is trivially true, no need to deal with the cross product.