Let $R$ denote the set of all real numbers. $B$ is any Bernstein set of $R$. 



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> Bernstein Set: A subset of the real line that meets every uncountable closed subset of the real line but that contains none of them. It's from wiki.
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We topologize $R$ now: the set $B$ is discrete and its complement has the usual topology. How to see the new topological space is Lindelof?



How to see such space is Lindelof?
Note that the open subsets of (what I will denote by) $\mathbb{R}\_B$ are of the form $U \cup A$ where $U \subseteq \mathbb{R}$ is open in the usual topology, and $A \subseteq B$ is arbitrary. 


Suppose that $\{ U\_i \cup A\_i : i \in I \}$ is an open cover of $\mathbb{R}\\_B$. Note that there is a countable $I\_0 \subseteq I$ such that $\bigcup\_{i \in I\_0} U\_i = \bigcup\_{i \in I} U\_i$. Next note that $\mathbb{R} \setminus \bigcup\_{i \in I} U\_i \subseteq B$ is closed (in $\mathbb{R}$) and is therefore countable, so there is a countable $I\_1 \subseteq I$ such that $\mathbb{R} \setminus \bigcup\_{i \in I} U\_i \subseteq \bigcup\_{i \in I\_1} A\_i$.


