When is discreteness equivalent to strong discreteness? 
Definition: say a set $D \subset X$, for some topological space $X$, is discrete if its induced topology is discrete.


Definition: say a set $D \subset X$, for some topological space $X$, is strongly discrete if there is a pairwise disjoint collection $\{O_x\}_{x \in D}$ such that $O_x$ is a neighbourhood of $x$.

In general, we can have topological spaces $X$ and $D \subset X$ such that $D$ is discrete but not strongly discrete. For instance, consider any finite subset of $\mathbb{N}$ with the cofinite topology.

Question: when are discreteness and strong discreteness equivalent?

I'm almost certain discreteness and strong discreteness are equivalent in $\mathbb{R}^n$, and generally in metric spaces, but I'm not sure. I'm less certain, but somewhat confident, that the equivalence should hold in Hausdorff spaces.
I'd be happy with an answer just proving the equivalence for $\mathbb{R}^n$, which is what I'm most interested in.
 A: Suppose $D$ is a discrete subset of a metric space $X$. For each point $x\in D$ there is a positive number $\varepsilon_x$ such that the ball of radius $\varepsilon_x$ centered at $x$ contains no other point of $D$. Let $O_x$ be the open ball of radius $\frac13\varepsilon_x$ centered at $x$. Then the sets $O_x$ ($x\in D$) are pairwise disjoint, which shows that $X$ is strongly discrete.
A: It’s not enough for $X$ to be Hausdorff. Let $X$ be $\Bbb R$ with the K-topology: if $\tau_E$ is the Euclidean topology on $\Bbb R$, the topology on $X$ is
$$\tau=\tau_E\cup\{U\setminus K:U\in\tau_E\}\,,$$
where $K=\left\{\frac1n:n\in\Bbb Z^+\right\}$. $X$ is Hausdorff but not regular. Let $D=K\cup\{0\}$. Then $X$ is a closed, discrete subset of $X$, but if $U_0$ is an open nbhd of $0$, and $U_n$ is an open nbhd of $\frac1n$ for each $n\in\Bbb Z^+$, $U\cap\bigcup_{n\in\Bbb Z^+}U_n\ne\varnothing$: no matter how you choose the nbhds $U_n$ for $n\in\Bbb Z^+$, any open nbhd of $0$ must meet all but finitely many of them.
In fact we can say much more. An old example due to R.H. Bing and known as Example H is $T_6$ (Hausdorff and perfectly normal) and therefore completely normal and hereditarily normal, but it is not collectionwise Hausdorff: it has a closed, discrete subset whose points cannot be separated by pairwise disjoint open sets. Thus, even rather strong separation axioms are not enough to ensure that discreteness implies strong discreteness.
Example H is rather complicated, and I’ll not describe it here; there is a full exposition in Dan Ma’s Topology Blog.
