Define topology on $\mathbb{R}$ as $\mathbb{R}\setminus\{1\}$,$\mathbb{R}\setminus\{2\}$, $\mathbb{R}\setminus\{1,2\}$, $\mathbb{R}$ are open sets in topology where $\mathbb{R}$ is real line.
Is $\mathbb{R}\setminus\{0\}$ Lindelöf in subspace topology?
Since topology contains finite number of open sets then space is compact, hence Lindelöf. But I cannot understand why each singleton in $\mathbb{R}\setminus\{0\}$ is open.
Indeed $X$ has a finite topology. So $X$ and all its subspaces are even compact, not just Lindelöf.
In this topology iit's not the case that all singletons in $\mathbb{R}\setminus\{0\}$ is open. If that is claimed, the topology must be different : maybe the included point topology wrt $0$?
Or maybe you mean $\mathbb{R}$ in the excluded point topology wrt $0$? The topology is $\{\mathbb{R}\} \cup \{O \subseteq \mathbb{R}: 0 \notin O\}$ and $\mathbb{R}$ is compact while all singletons of $A=\mathbb{R}\setminus \{0\}$ are open ( as $\{x\}$ is open if $x\neq 0$) and so $A$ is not Lindelöf in the subspace topology.