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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.

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    For what it's worth, the existence of such a $\tau'$ is equivalent to the existence of a countable set of continuous functions $X\to [0,1]$ that separate points of $X$. – 2017-02-26
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    @EricWofsey why just points, not points and closed sets as well? – 2017-03-08
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    @HennoBrandsma: Because you only need $\tau'\subseteq\tau$, not $\tau'=\tau$. – 2017-03-08
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    @EricWofsey So continuous in the $\tau$-topology on $X$. – 2017-03-08
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    @HennoBrandsma: Yes, I was stating a condition entirely in terms of the original topology $\tau$. – 2017-03-08
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    @EricWofsey In that case the answer is clear, there are $T_{2.5}$ Spaces with only constant real-valued functions. So no such $\tau'$ then exists. – 2017-03-08
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    @HennoBrandsma Could you point me to a second-countable example? – 2017-03-08
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    @HennoBrandsma: But are there such spaces that are second-countable? – 2017-03-08
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    @EricWofsey Arens square .(See counterexamples) This has points that cannot be separated by continuous functions. – 2017-03-08
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    @EricWofsey It does have a coarser metric topology though. So your condition is not necessary, it seems. – 2017-03-08
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    From the same source ,Roy's lattice space. – 2017-03-09

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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 $02/3$ whenever $3/4

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.