Proof:

The quotient topology on $Y$ induced by $f$ is stronger than the given topology. To obtain the reverse inclusion, suppose that $f$ is a continuous open mapping and $A \subset Y$ is open with respect to the quotient topology on $Y$ induced by $f$ which means $f^{-1}(A)$ is open in $X$ whereby $f(f^{-1}(A))$ is open in $Y$ in the given topology since $f$ is an open mapping. But since $f$ is surjective, $f(f^{-1}(A)) = A$ and so we conclude $A$ is open in the given topology as well.

Let us now turn to a continuous, closed surjective map $f : X \longrightarrow Y$. Again we merely have to show that the given topology on $Y$ is stronger than the quotient topology since the reverse inclusion is trivial. So let $A$ be an open set in $Y$ with respect to the quotient topology induced by $f$. By definition $f^{-1}(A)$ is open in $X$, or in other words $X - f^{-1}(A)$ is closed in $X$. Since $f$ is closed, $f(X-f^{-1}(A)) = Y - A$ is closed in $Y$ with respect to the given topology on $Y$. That is to say $A$ is open with respect to the given topology on $Y$.