Let $X$ be a topological space. $A\subseteq X$ is Lindelöf $\iff$ Every cover of $A$ by open subsets of $X$ has a countable subcover.
My attempt:
Let $(U_i)_{i \in I}$ be a cover for $A$ by open subsets of $X$. So $A\subseteq \bigcup_{i\in I}U_i$. Then $A\subseteq A \cap $ $\bigcup_{i\in I}U_i$ $=$ $\bigcup_{i\in I } A\cap U_i$ ,which is a cover for $A$ by sets open relative to $A$. By assumption, there exists a countable subcover, indexed by $J\subseteq \mathbb{N}$ such that $A\subseteq \bigcup_{j\in J}A\cap U_j$. This implies that $A\subseteq \bigcup_{j\in J}U_j$, which means that every cover of $A$ by open subsets of $X$ has a countable subcover. For the converse, let $(U_i)_{i\in I}$ be sets open relative to $A$ that cover $A$. So $A\subseteq \bigcup_{i\in I}U_i$, where each $U_i \in \tau_A$. Hence, $U_i=U_i' \cap A$ for some $U_{i}' \in \tau_X$. Hence $A\subseteq$ $\bigcup_{i\in I}(U_i' \cap A)$ $=$ $A \cap \bigcup_{i\in I}U_i'$ $\implies$ $A\subseteq$ $\bigcup_{i\in I}U_i'$, and so by assumption there exists an enumeration $U_1',U_2',U_3'...$ by an index set $J\subseteq \mathbb{N}$ such that their union covers $A$, hence $A\subseteq A\cap (U_1' \cup U_2' \cup U_3' \ldots)$ $=$ $\bigcup_{j \in J} (A\cap U_j')$, and so $A$ is Lindelöf.
Is it correct?
$J$ need not be a subset of $\Bbb N$. It's just a countable subset of $I$, and just say so.
Moreover, we can state $$A = A \cap \bigcup_{i \in I} U_i$$ etc. with equality, so that $A$ is exactly covered by the relatively open cover $\{U_i \cap A: i \in I\}$, which has a countable subcover $\{U_i \cap A: i \in J\}$ where $J \subseteq I$ countable, and then $$A = \bigcup_{i \in J} (U_i \cap A) = A \cap \bigcup_{i \in J} U_i$$ implying $A \subseteq \bigcup_{i \in J} U_i$ as required.
The other direction has an equality instead of an inclusion too in $A = \bigcup_{i \in I} U_i$ as well, when the $U_i$ are open in $A$ (which implies already they are all a subset of $A$). Same remark about $J$ just being countable subset of $I$, not of $\Bbb N$.