This is the definition of a Lindelöf space:
A space for which every open covering contains a countable subcovering is a Lindelöf space.
In the proof that $\mathbb{R}_l$ is Lindelöf, they have this statement:
It will suffice to show that every open covering of $\mathbb{R}_l$ by basis elements contains a countable subcovering of $\mathbb{R}_l$.
But I can not see why:
My attempt at proof:
Let $\{O_a\}$ be an open covering, set $A=\cup O_a$. A must be open. For each $x \in A$, there is a basis element such that $x \in B_x\in A$. But the problem is that for $x \in A^c$, $A^c$ is closed, so how do we handle these elements? I can not conclude that we have a countable covering of A, since even though we have basis elements covering A, these elements may not cover the entire space $\mathbb{R}_l$?
Given an open covering $\{O_a\}$ we can write each $O_a$ as union of basis elements, $O_a=\bigcup_{i\in I_a}B_{\langle a,i\rangle}$. Then $A$ is also covered by all those $B_{\langle a,i\rangle}$. If we can find a countable subcover $\{\,B_{\langle a_n,i_n\rangle}\mid n\in\Bbb N\,\}$ then certainly $\bigcup_{n\in\Bbb N}O_{a_n}$ covers $A$.