[2]$\Rightarrow$ [1]. It is easy to check that each discrete subset $D’$ of a perfect space $P$ is nowhere dense in $P$. Now suppose that $D$ is a countable union of discrete subsets of $X$ and $D$ contains a non-empty perfect subset $P$ of $X$. By Theorem 4.3.11 from [Eng], $P$ is completely metrizable. By Theorem 2.4 from [Eng], $P$ is a Baire space, therefore it cannot be a countable union of its discrete subsets.
[1]$\Rightarrow$ [2]. I googled a little and found articles [Sto] and [Kou]. In the beginning of the paper [Kou] it is written the following: “A classical theorem of Suslin states that every analytic subset of a Polish (separable complete metric) space is either countable, or contains a copy of the Cantor set. A generalization to the non-separable case has been obtained by El’kin [1]: every absolutely analytic space (i.e. homeomorphic to a Suslin subset of some complete metric space) is either $\sigma$-discrete, or contains a copy of the Cantor set. This theorem had previously been proved by Stone [Sto] for absolutely Borel spaces”.
Concerning descriptive generalizations for Polish spaces, I recall that book [Kech] contains The Perfect Sets Theorems for Borel (13.6) (by Alexandrov and Hausdorff), Analytic (14.13) (by Souslin), Co-Analytic (32.2) and Projective (38.17) (by Davis) Sets: each such subset of a Polish space either is countable or else it contains a copy of Cantor set.
It seems that these results can be trivially generalized to locally separable metrizable spaces (by [Eng, Ex. 4.4.F.C] each such space is a disjoint sum of separable spaces) and to $\sigma$-discrete subsets instead of countable.
I remark that if we drop descriptive conditions for the set $D$ (like Borelness) then [1] does not imply [2]. Let $D$ be a Bernstein subset (see [Cic]) of the real line $\mathbb R$ (for non-separability we can take as $X$ a disjoint sum of uncountably many copies of $\mathbb R$). Each discrete subset of a hereditarily separable space (in particular, of a second countable space) is countable. Therefore each countable union of discrete subsets of $\mathbb R$ is countable. But if $D$ were countable, then $\mathbb R\setminus D$ will be a completely metrizable space as a non-empty $G_\delta$ subset of a completely metrizable space by Theorem 4.3.23 from [Eng]. Therefore $\mathbb R\setminus D$ will contain a copy $C$ of a Cantor set, [Cic] a contradiction.
References
[Cic] Jacek Cichoń. On Bernstein Sets.
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
[HC] R. C. Haworth, R. C. McCoy, Baire spaces, Warszawa, Panstwowe Wydawnictwo Naukowe, 1977.
[Kech] Alexander S. Kechris. Classical Descriptive Set Theory , Springer, 1995.
[Kou] George Koumoullis. Cantor sets in Prokhorov spaces.
[Sto] A. H. Stone. Kernel constructions and Borel sets.
