Here goes a simpler proof (in my opinion). As in the previous one, it doesn't need to be assumed that the space is regular, the desired follows just by assuming Hausdorff. Notice that, in spaces that are at least $T_1$, $A$ is a closed discrete subset iff $\{\{x\}: x \in A\}$ is locally finite. So, take the very same increasing sequence of compact subsets, say $\{K_n: n < \omega\}$, such that $X = \bigcup\limits_{n < \omega} K_n$. Let $\langle x\rangle_0$ be a finite subset of $K_0$ such that $N[\langle x\rangle_0]$ covers $K_0$ (where, of course, $N[\langle x \rangle_0]$ is a short for $N[\textrm{im}(\langle x \rangle_0)]$). Then $\{N(x): x \in K_1 \setminus N[\langle x\rangle_0]\}$ covers the compact set $K_1 \setminus N[\langle x\rangle_0]$ - and this is the point where we use Hausdorffness, to be sure that $K_1$ is a closed set, and therefore $K_1 \setminus N[\langle x\rangle_0]$ is a closed subset of a compact set -, and so there is a finite sequence $\langle x \rangle_1$ of elements of $K_1 \setminus N[\langle x\rangle_0]$ such that $N[\langle x \rangle_0\, ^\frown \langle x \rangle_1]$ covers $K_1$. Proceeding inductively, if $j > 1$ there is a finite sequence $\langle x \rangle_j$ of elements of $K_j \setminus N[\langle x \rangle_0\, ^\frown \langle x \rangle_1\,\frown \ldots \frown \langle x \rangle_{j -1 }]$ such that $N[\langle x \rangle_0\, ^\frown \langle x \rangle_1\,\frown \ldots \frown \langle x \rangle_j]$ covers $K_j$. Now, with all finite sequences $\langle x \rangle_n$ constructed, let $x$ be the infinite sequence $x = \langle x \rangle_0\,\frown\langle x \rangle_1 \frown \ldots \frown \langle x \rangle_n \frown \ldots$ and take $F = im(x)$. It is clear that $N[F] = X$. We can check that $F$ is closed and discrete at once: it suffices to verify the local finiteness of the family of singletons. Indeed: it $t \in X$, let $m = \textrm{min}\{j: t \in K_j\}.$ By construction, $V = N[\langle x \rangle_0\, ^\frown \langle x \rangle_1\,\frown \ldots \frown \langle x \rangle_m]$ is an open neighbourhood of $t$ which contains only finitely many elements of $F$.