 Solution

For any $c\in C$, the surjectivity of $g\circ f$ supplies $a\in A$ with $g(f(a))=c$. Therefore $f(a)\in B$ is a preimage of $c$ under $g$, proving surjectivity of $g$.

For the requested counterexample, take $A=C=\{0\}$, $B=\{0,1\}$, $f(0)=0$, and $g(0)=g(1)=0$. The function composition is the identity bijection of the singleton, but $f$ is not surjective and $g$ is not injective.