$T_{2.5}$ topology without coarser metric topology Let $(X,\tau)$ be a second-countable $T_{2.5}$ space, where with $T_{2.5}$ I mean that any distinct points are separated by closed neighborhoods. Does there have to be some metrizable second-countable $\tau' \subseteq \tau$?
The typical examples of $T_{2.5}$ spaces that are not metrizable seem to be constructed by adding additional open sets to some metrizable topology, so I would be interested in a potential example of a space which is constructed differently -- or maybe a proof that we can always find a coarser metrizable second-countable topology.
 A: As pointed out by Henno Brandsma in the comments, an example is the "Arens square" as modified by Brian Scott in his answer to this question.  This space $(X,\tau)$ is $T_{2.5}$ (thanks to Brian Scott's modification), and is second-countable since it has only countably many points and it is clearly first-countable.
However, there is no continuous function $f:X\to [0,1]$ such that $f(0,0)=0$ and $f(1,0)=1$.  Indeed, given such a function $f$, there would be $\epsilon>0$ such that $f(x,y)<1/3$ whenever $0<x<1/4$ and $0<y<\epsilon$ and $f(x,y)>2/3$ whenever $3/4<x<1$ and $0<y<\epsilon$.  It then follows by continuity that $f(1/4,y)\leq 1/3$ and $f(3/4,y)\geq 2/3$ for $0<y\leq \epsilon$.  Now fix $q\in Q_{1/2}$ such that $q<\epsilon$.  Note that every closed neighborhood of $(1/2,q)$ contains a point of the form $(1/4,y)$ and also a point of the form $(3/4,y')$ for $y,y'<\epsilon$.  It follows that $f(1/2,q)$ must be both $\leq 1/3$ and $\geq 2/3$, which is a contradiction.
If a coarser metrizable topology $\tau'\subseteq\tau$ existed, then there would exist such a function $f$ that is continuous with respect to $\tau'$, and hence also with respect to $\tau$.  Thus no such $\tau'$ exists.
