Let $X$ and $Y$ be two simplicial sets. Then there is a natural continuous map $\lvert X \times Y \rvert \to \lvert X \rvert \times \lvert Y \rvert$, where the right-hand side is given the product topology. I've often seen the assertion in the literature that this map is not necessarily a homeomorphism, hence the need to work in some more convenient category.
What is an example for which this map fails to be a homeomorphism? I suspect that the simplicial sets involved in such a counterexample will be quite "large", but I haven't seen enough cases of strange geometric realizations to come up with the counterexample on my own.
The following counterexample is apparently due to Dowker and is taken from the appendix of Hatcher's Algebraic Topology (p. 524-5). Let $X=\bigvee_{f\in S} I_f$ be a wedge sum of continuum many $1$-simplices $I_f$, indexed by the set $S$ of all functions $\mathbb{N}\to(0,1]$. Let $Y=\bigvee_{n\in\mathbb{N}} I_n$ be a wedge sum of countably infinitely many $1$-simplices $I_n$, indexed by the natural numbers. Given $f\in S$ and $n\in\mathbb{N}$, define $p_{fn}=(f(n),f(n))\in |I_f\times I_n|\subset|X\times Y|$, where we identify $|I_f\times I_n|$ with $[0,1]^2$, where $0$ corresponds to the wedge point in both coordinates. Let $P\subset|X\times Y|$ denote the set of these points $p_{fn}=(f(n),f(n))$ for all $f$ and $n$. Then $P$ is closed in $|X\times Y|$ (its intersection with any simplex consists of at most one point).
However, $P$ is not closed as a subset of $|X|\times |Y|$ with the product topology. Indeed, letting $0$ denote the wedge point of both $|X|$ and $|Y|$, I claim that $(0,0)$ is in the closure of $P$ in $|X|\times |Y|$. To prove this, let $U$ be a neighborhood of $0$ in $|X|$ and $V$ be a neighborhood of $0$ in $|Y|$. For each $n\in\mathbb{N}$, there is some $t_n\in(0,1]$ such that $V\cap |I_n|$ contains $[0,t_n)$, identifying $|I_n|$ with $[0,1]$. We may assume the sequence $(t_n)$ converges to $0$ (if it doesn't, just make the numbers $t_n$ smaller). Now define $f:\mathbb{N}\to(0,1]$ by $f(n)=t_n/2$. Then $U\cap |I_f|$ contains $[0,s)$ for some $s>0$, identifying $|I_f|$ with $[0,1]$. Now choose $n$ sufficiently large so that $f(n)<s$. We then find that the point $p_{fn}=(f(n),f(n))\in |I_f\times I_n|$ is contained in $U\times V$. That is, $U\times V$ intersects $P$. Since $U$ and $V$ were arbitrary neighborhoods of $0$, this means $(0,0)$ is in the closure of $P$ with respect to the product topology.