Let $(X,\tau)$ be a nonmetrizable topological space which is not firstcountable and let $\emptyset \neq Y\subset X$ be a proper dense subset. Is it possible for $(Y,\tau_Y)$ (where $\tau_Y$ is the relativisation of $\tau$ to $Y$) be metrizable?

$\begingroup$ Since second countability wouldn't work I edited your question to require first countability. $\endgroup$ – AIM_BLB Jan 14 at 10:11

$\begingroup$ You're right, that does indeed make sense. $\endgroup$ – AnnieTheKatsu Jan 14 at 10:12

$\begingroup$ Did you mean to ask whether it's possible for $Y$ to be metrizable? As you've phrased it, the answer to your question seems obvious: of course it's possible for $Y$ to be nonmetrizable, e.g., if we take $Y = X$. $\endgroup$ – Will Brian Jan 14 at 13:52

$\begingroup$ @WillBrian Indeed, I had made an English hickup. $\endgroup$ – AnnieTheKatsu Jan 14 at 13:53

4$\begingroup$ Another class of examples is to let $Y$ be any noncompact metrizable space, and $X = \beta Y$ its StoneČech compactification. $\endgroup$ – Nate Eldredge Jan 14 at 18:03
Yes: the order topology on $\omega_1+1$ (the first uncountable successor ordinal) is an example. It is not first countable, because it's "top" point $\omega_1$ has no countable neighborhood base. But the set of all isolated points of this space is dense in it, and the relative topology on this set is discrete (hence metrizable).
On the other hand, let me point out that if $x \in X$ has no countable neighborhood base, and if $X$ is $T_3$, then $x$ will still fail to have a countable neighborhood base in any dense $Y \subseteq X$ with $x \in Y$. Therefore there are plenty of spaces $X$ for which the answer to your question would be negative. For example, no dense subset of $[0,1]^{\kappa}$ is metrizable (for uncountable $\kappa$), because every point of this space witnesses the fact that the space is not firstcountable, and this will continue to be true in any dense subspace.

$\begingroup$ Do you have a reference to the second point? $\endgroup$ – AnnieTheKatsu Jan 14 at 14:34

$\begingroup$ I needed a version of this lemma one time in an old paper: see lemma 4.5 in wrbrian.files.wordpress.com/2012/01/neight2.pdf. But I'm sure I wasn't the first person to notice this fact  if you're willing to look hard enough, I wouldn't be surprised if you'd find it in Engelking. $\endgroup$ – Will Brian Jan 14 at 14:39

$\begingroup$ Also, for a slightly smaller example, the order topology on $\omega_1$ is not metrizable. $\endgroup$ – François G. Dorais♦ Jan 14 at 21:26

$\begingroup$ The onepoint compactification of an uncountable discrete space is an even simpler example. More generally, every nonfirst countable weakly compact subspace of a Banach space is an example. $\endgroup$ – Santi Spadaro Jan 14 at 22:17

1$\begingroup$ @FrançoisG.Dorais: The OP asks that the original space not be first countable. While $\omega_1$ is not metrizable, it is first countable. $\endgroup$ – Will Brian Jan 14 at 22:40