A space $X$ is said to be strongly starLindelöf if for every open cover $\mathcal U$ of $X$ there exists a countable subset $A$ of $X$ such that $St(A,\mathcal U)=X$.
A space $X$ has discrete countable chain condition (DCCC) if every discrete family of nonempty open sets is countable.
Every strongly starLindelöf $T_1$space is DCCC. Let $\mathcal{U}$ be a discrete family of nonempty open sets in the space $X$, and pick a point $x_U\in U$ for each $U\in\mathcal{U}$. The set $F=\{x_U:U\in\mathcal{U}\}$ is closed (by discreteness of $\mathcal{U}$). Then $\mathcal{V}=\mathcal{U}\cup\{X\setminus F\}$ is an open cover and for every point $x$ in $X$ the star $\operatorname{St}(x,\mathcal{V})$ meets $F$ in at most one point. By strong starLindelöfness $F$ is countable and hence so is $\mathcal{U}$.

$\begingroup$ @Hart: Why $F$ is closed? Can you expalin it? $\endgroup$– Nur AlamSep 2 '21 at 7:54

2$\begingroup$ I tend to assume that the spaces I work with are Hausdorff (or at least $T_1$). In that case if $x\in X\setminus F$ then $x$ has a neighbourhood $O$ that meets at most one $U\in\mathcal{U}$. Deleting that one point from $O$ leaves a neighbourhood disjoint from $F$. I added $T_1$ to my answer. $\endgroup$– KP HartSep 2 '21 at 8:14

$\begingroup$ @Hart: Actually we do not assume any separation axiom. $\endgroup$– Nur AlamSep 2 '21 at 8:41