Well, I've found a reference here: http://home.imf.au.dk/marcel/gentop/noter.pdf. Exercise 4.4.7 in those notes describes an example, which is also (I think) referred to here: Products of quotient topology same as quotient of product topology. I'll wave my hands at it a little bit - feel free to correct me if things seem wrong.
Let $X$ be the set of positive rational numbers, and define an equivalence relation on it by declaring any two positive integers to be equivalent. Call that $\sim_{X}$. Let $Y = \mathbb{Q}$, and define the trivial equivalence relation on it, so we identify no points. Then the claim is that $X \times Y/\sim$ is not homeomorphic to $X/\sim_{X} \times Y$.
Define $$ U_{n} = \left\{ (x,y) \in X \times Y \,| \, |x-n| < \textrm{min}(|y - \frac{\sqrt{2}}{n}|,1/2) \right\}.$$ Then set $U = \cup_{n \geq 1} U_{n}$. So each $U_{n}$ is a set around $(n,0)$ in $X \times Y$, where $y$ can be whatever it wants, $x$ is always between $n-1/2$ and $n+1/2$, and if $y$ is close to $\sqrt{2}/n$, then $x$ has to be even closer to $n$. I picture it as sort of an infinitely tall hourglass, where it pinches in near $y = \sqrt{2}/n$.
Then if $q$ is the quotient map from $X \times Y$ to $X \times Y/\sim$, $q^{-1}(q(U)) = U$, so that $q(U)$ is open. This is relatively easy to see - $q$ is only non-injective at points of the form $(m,y)$, where $m$ is an integer. But if $(m,y) \in U$, then every point $(n,y) \in U_{n}$, because $y - \frac{\sqrt{2}}{n}$ can never be $0$, so that $x=n$ is always allowed in $U_{n}$.
Now, if $\epsilon > \sqrt{2}/n$, then there are no $a <n < b$ so that the cylinder $(a,b) \times (-\epsilon, \epsilon) \cap X \times Y$ is contained in $U_{n}$. Essentially, this is because $(a,b) \times (-\epsilon,\epsilon)$ contains the "pinch" of the hourglass, which gets arbitrarily narrow, so no interval $(a,b)$ is narrower than it.
Now, we look at $f(q(U))$. We claim it is not open because it contains no neighborhood of $([1],0)$. Let $\epsilon >0$ be given and assume that $\epsilon < n/ \sqrt{2}$. (Just look at a large enough $n$ to make that true.) If $f(q(U)$ were open, we could find a neighborhood of $([1],0)$ contained in $f(q(U))$ whose $y$-coordinates were all in $(-\epsilon,\epsilon)$. But the previous paragraph essentially proves that can't happen.
Intuitively, $q(U)$ is open because no matter what $n$ is, $q(U)$ contains a neighborhood of $(n,y)$ - the hourglass has SOME width everywhere, even though it's arbitrarily small. But, if you fix $y=0$ and try and put a neighborhood around every $(n,0)$ all at once (i.e. put a neighborhood around $([1],y)$), then for large enough $n$, your small neighborhood will extend around the pinch of the hourglass for that particular $n$.
