Can we state the following?
Let $X$ be a topological space with a basis $\mathscr{B}$ for the topology of $X$. If every covering of $X$ by basis sets from $\mathscr{B}$ has a countable subcollection that also covers $X$, then $X$ is a Lindelof space.
Here is the start of my (proposed) proof:
Let $\mathscr{A}$ be any open covering of $X$. For each $A \in \mathscr{A}$, let $\mathscr{B}_A$ be a collection of sets from basis $\mathscr{B}$ whose union is $A$. Note that every open set in a topological space is a union of some basis sets. Let $\mathscr{B}_\mathscr{A}$ denote the union of the subcollections $\mathscr{B}_A$ as $A$ ranges over $\mathscr{A}$, that is, let us put $$ \mathscr{B}_\mathscr{A} \colon= \bigcup_{A \in \mathscr{A} } \mathscr{B}_A. \tag{Definition 0} $$ Then as we have $$ X = \bigcup_{A \in \mathscr{A} } A \tag{0} $$ and, for each $A \in \mathscr{A}$, we have $$ A = \bigcup_{B \in \mathscr{B}_A} B, \tag{1} $$ so we also have $$ \begin{align} X &= \bigcup_{A \in \mathscr{A} } A \qquad \mbox{[ using (0) above ] } \\ &= \bigcup_{A \in \mathscr{A} } \left( \bigcup_{B \in \mathscr{B}_A} B \right) \qquad \mbox{[ using (1) above ] } \\ &= \mbox{[ Am I right? ]} \bigcup_{B \in \mathscr{B}_\mathscr{A} } B. \tag{2} \end{align} $$ Thus the collection $\mathscr{B}_\mathscr{A}$ is a covering of $X$ by (some) basis sets. Therefore there exists a countable subcollection $\mathscr{B}_\mathscr{A}^*$ of $\mathscr{B}_\mathscr{A}$ that also covers $X$, by our hypothesis. For each $A \in \mathscr{A}$, let us put $$ \mathscr{B}_A^* \colon= \mathscr{B}_A \bigcap \mathscr{B}_{\mathscr{A}}^*. \tag{Definition 1} $$
Is the argument so far correct? If so, then how to proceed from here?
Or, is there some other (easy and elementary enough) approach to proving this result?
$\newcommand{\B}{\mathscr{B}}\newcommand{\A}{\mathscr{A}}$This is good. Take the countable subcollection $\B^\ast_\A$; we can choose a countable subset $\A^\ast\subseteq\A$ such that every $G\in\B^\ast_\A$ is contained in some $A_G\in\A^\ast$. Then: $$X=\bigcup\A^\ast$$Follows; $\A^\ast$ is a countable subcover of $\A$, as desired.