Proof for $\mathbb{R}_{l}$ is Lindelof

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In Munkers there is a proof that $\mathbb{R}_{l}$ is Lindelof. $\mathcal{A}$ be the collection of basis elements of the form $\{ [a_{\alpha},b_{\alpha}) : \alpha \in J \}$ be a covering of $\mathbb{R}$. Let $C= \bigcup_{\alpha \in J} (a_{\alpha},b_{\alpha})$ which is a subset of $\mathbb{R}$.

Then it is shown that $\mathbb{R} - C$ is countable. Then construct a sub collection of basis elements in $\mathcal{A}$ using this $\mathbb{R} -C$ which is countable and cover $\mathbb{R}- C$. In next step we will topologize $C$ as a subspace of $\mathbb{R}$. Then using second countability of $C$ we will construct another countable subcollection of $\mathcal{A}$ which cover $C$. Both the collections together forms a countable subcollection of $\mathcal{A}$ which cover $\mathbb{R}_{l}$.

My doubt is in the construction of $C$. Is there a chance that $C$ differ from $\mathbb{R}$ ? Even it is equal to $\mathbb{R}$ we are going to get first countable subcollection is going to be non empty. Can we find a particular covering for $\mathbb{R}_{l}$ by basis elements of the form $\{ [a_{\alpha},b_{\alpha}) \}$ such that $\mathbb{R} -C$ is non empty ? Provide an example.

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Will this particular case is sufficient ? $\mathcal{A}= \{[n,n+1) \}$ where $n$ is an integer. Then even though $\mathcal{A}$ covers $\mathbb{R}_{l}$, $C$ does not contain any of the integers.