As in the previous exercise we want to cover $X$ with a countable collection of open subspaces such that each one has a countable basis. For $x$ take open metrizable $U$ containing $x$ , find open $V$ such that $x\in V\subseteq\overline{V}\subseteq U$ . $\overline{V}$ is a closed subspace of a Lindelöf space, i.e. Lindelöf. It is also metrizable, therefore, being Lindelöf is equivalent to being second-countable. Hence, $V$ is an open second-countable (in the subspace topology) neighborhood of $V$ . Now cover $X$ with such neighborhoods, find a countable subcovering, and prove that the countable union of countable bases of all open sets in the subcovering is a countable basis of $X$ . Therefore, $X$ is regular and second-countable. Hence, metrizable. We used the regularity twice: to find a neighborhood with the closure within a given neighborhood, and to argue that the space being regular and second-countable is metrizable. If the space is Hausdorff, Lindelöf and locally metrizable then it is not necessarily metrizable. To find a counterexample we need a space which is Hausdorff but not regular. $\mathbb{R}_K$ is such a space. It is also Lindelöf but not metrizable. The only question is whether it is locally metrizable. The subspace topology on $(-\infty,0)$ and $(0,+\infty)$ is the same as the standard sub-topology, therefore, metrizable. The only question is now whether 0 has a metrizable neighborhood. But $\mathbb{R}-K$ is its neighborhood with the same topology as the standard one, therefore, it is metrizable.