Statement: $g:A \to B$ and $f:B \to C$ are injective functions. Then $f \circ g:A \to C$ is injective.
Attempt of proof:
Suppose by contradiction that $f \circ g$ is not injective. Then there exist $x$ and $y$, $ x \neq y$, such that $f \circ g(x) = f\circ g(y)$. Then, by definition of composition, $f(g(x)) = f(g(y))$. Since $g$ is injective, $f(x) = f(y)$. Since $f$ is injective, $x = y$. But that is a contradiction, since $x \neq y$.
I'm not sure if every step I'm doing here is alright, specialy that if $f \circ g$ isn't injective then there exist $x$ and $y$, $x \neq y$. Why there must be two points in $A$?