Let $(X,\tau)$ be a non-metrizable topological space which is not first-countable 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?



Restriction of non-metrizable topology to dense subset is non-metrizable
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 first-countable, and this will continue to be true in any dense subspace.


