= Solution
Since $D$ is a <discrete subset> in the <subspace topology>, for each $d\in D$ choose an <open set> $U_d\subseteq Z$ with $U_d\cap D=\{d\}$. Closedness of $D$ makes $Z\setminus D$ <open>. The family
$$
\{Z\setminus D\}\cup\{U_d:d\in D\}
$$
is therefore an <open cover> of $Z$. A <countable> subcover exists by the <Lindelöf> property. Each point $d$ can be covered only by its own $U_d$ among this family: the <complement> misses all of $D$, and every other $U_e$ meets $D$ only at $e$. Hence the <countable> subcover must include $U_d$ for every $d\in D$. This forces
$$
\boxed{D\text{ is countable}.}
$$
Both closedness and discreteness matter: they provide the <open> <complement> and the single-point traces, respectively.
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