If $A \sim B$, it means there is a one-to-one and onto function from $A$ to $B$. If $A$ and $B$ are sets, then we will say that $B$ dominates $A$, and write $A \precsim B$, if there is a function $f : A \rightarrow B$ that is one-to-one. If $A \precsim B$ and $A \nsim B$, then we say that $B$ strictly dominates $A$, and write $A \prec B$.
(b) For all sets $A$, $B$, and $C$, if $A \prec B$ and $B \prec C$, then $A \prec C$.
Suppose $A \prec B$ and $B \prec C$. Then, there are functions $f : A \rightarrow B$ and $g : B \rightarrow C$ that are one-to-one but not onto. Let $h : A \rightarrow C$ be defined as $h = g \circ f$. Suppose $h (a_1) = h (a_2)$. Then, $g (f (a_1)) = h (a_1) = h (a_2) = g (f (a_2))$. Since $g$ is one-to-one, $f (a_1) = f (a_2)$. Since $f$ is one-to-one, $a_1 = a_2$. Thus, $h$ is one-to-one, and $A \precsim C$.
So far so good. But, I can't prove that $A \nsim C$. How would I prove that $A \nsim C$ and complete the proof?
I cannot find a proof without using Schröder–Bernstein theorem or something like that.
Indeed, you need to prove that if $A \precsim B$ and $B \precsim A$ then $A\sim A$. If you know this result, then suppose $A\sim C$. We have $B\prec C$ and $C\sim A$; hence $B\precsim A$. Thus $A\sim B$, contradiction.