The one-point compactification of the discrete real line is compact and Lindelöf but is neither first countable nor separable

Give $D=\mathbb{R}$ the discrete topology and form $D^{\ast }=D\cup \{\mathrm{\infty }\}$. The one-point compactification theorem makes $D^{\ast }$ compact, hence Lindelöf, and every point of $D$ remains isolated.

A neighbourhood of $\mathrm{\infty }$ has finite complement in $D$, since compact subsets of a discrete space are finite. If $(N_n)$ were a countable local base at $\mathrm{\infty }$, put $F_n=D\setminus N_n$. For every $x\in D$, the neighbourhood $D^{\ast }\setminus \{x\}$ would contain some $N_n$, so $D={\bigcup }_n{F}_n$. Each finite subset of $\mathbb{R}$ has a canonical increasing enumeration; these enumerations and countability of $\mathbb{N}\times \mathbb{N}$ make the displayed union countable, contradicting uncountability of $\mathbb{R}$. Thus $D^{\ast }$ is not first countable. Finally every dense subset must meet the open singleton $\{x\}$ for every $x\in D$, so it contains all of $D$ and cannot be countable; hence $D^{\ast }$ is not separable.