Proof:

If $g$ is continuous it is trivial that $g\circ f$ is continuous. Conversely suppose that $g\circ f$ is continuous. Let $A$ be any open set in $Z$ so that $(g\circ f)^{-1}(A)$ is open in $X$. Thus $f^{-1}(g^{-1}(A))$ is open in $X$. Invoking the definition of the quotient topology, we see that $g^{-1}(A)$ must be open in $Y$ which means $g$ is continuous.

Before illustrating the use of the universal property of quotients we discuss the following issue. Suppose that $X$ and $Y$ are topological spaces and $f : X \longrightarrow Y$ is a given continuous map, then the quotient topology on $Y$ induced by $f$ is weaker than the given topology on $Y$. When would the given topology on $Y$ be equal to the quotient topology?