Let $\pi_1 \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ be projection on the first coordinate. Let $A$ be the subspace of $\mathbb{R} \times \mathbb{R}$ consisting of all points $x \times y$ for which either $x \geq 0$ or $y = 0$ (or both); let $q \colon A \to \mathbb{R}$ be obtained by restricting $\pi_1$.
Then $q$ is a surjective map.
How to show that $q$ is a quotient map that is neither open nor closed?
Best Answer
To show that $q$ is a quotient map, you should probably just show directly that $q$ sends saturated open sets to open sets. Note that the saturated open sets are precisely those open sets $U$ such that if $(x,y)\in U$ with $x \ge 0$, then $\{x\} \times \Bbb R \subset U$.
Note that projection maps in general aren't closed: If you can figure out an example to show that $\pi_1$ isn't closed, then that technique will probably work for $q$ as well. One idea might be to choose a countable, non-closed set in $\Bbb R$, and to express it as a projection of a countable subspace of $\Bbb R^2$ consisting of isolated points.
Projection maps are open (by the definition of the product topology), so you are going to need to use something special about $A$ to find a suitable counterexample. Intuitively, the place where $q$ behaves badly is the line $x = 0$, since to the left $q$ acts as the identity and to the right $q$ acts as a projection. A reasonable open subspace whose image isn't open is therefore going to need to intersect $x=0$. Try an example and see if it works!