A locally metrizable, Lindelöf Hausdorff space that is not metrizable I am looking for an example of locally metrizable, Lindelöf Hausdorff space that is not metrizable.
I've proved that if such space is regular, then it is metrizable. The proof relies on the ability to find Lindelöf neighborhoods inside metrizable ones, so any counterexample must be non-hereditarily Lindelöf. Unfortunately, the only few examples of such spaces I'm familiar with fail to satisfy some conditions: the particular point topology on an uncountable set is not even T1; uncountable products of compact sets are not first countable; the unit square with lexicographic order is regular.
 A: An example of such a space is the K-topology on $\mathbb{R}$. Setting $K = \{ \frac{1}{n} : n \geq 1 \}$, the open sets are exactly the sets of the form $U \setminus A$ there $U \subseteq \mathbb{R}$ is open in the usual metric/order topology, and $A \subseteq K$.


*

*Since the topology is finer than the usual topology, it is clearly Hausdorff.

*It is not regular, since $0$ and the closed set $K$ cannot be separated by disjoint open sets. Therefore the space is not metrizable.

*It  is Lindelöf since given any open cover $\{ U_i \setminus A_i : i \in I \}$ you first take a countable $I_0 \subseteq I$ so that $\bigcup_{i \in I_0} U_i = \mathbb{R}$, and note that $\mathbb{R} \setminus \bigcup_{i \in I_0} ( U_i \setminus A_I ) \subseteq K$ is countable, so we need only add countably many more sets to cover $\mathbb{R}$.  (This idea can be modified to show that the space is hereditarily Lindelöf.)

*It is locally metrizable because every point has an open neighbourhood whose subspace topology with respect to the K-topology coincides with the subspace topology with respect to the usual topology. For $x \neq 0$ just take an open interval containing $x$ whose intersection with $K$ is finite.  For $x = 0$ consider $( -1  , 1 ) \setminus K$.
