 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 familyis 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 forcesBoth closedness and discreteness matter: they provide the open complement and the single-point traces, respectively.