Is this true? If $X$ has no isolated points and $D$ is dense, then every non-empty open of $X$ intersects $D$ infinitely many times? 
Proposition. Let $X$ denote a topological space and $D$ denote a dense subset thereof. Then for all open $U \subseteq X$, either $U$ is empty, or $|U \cap D| \geq 1.$

However, in $\mathbb{R}$ and $\mathbb{Q}$, we can say a lot more:

Proposition. Let $X=\mathbb{R}$ or $\mathbb{Q}$. Then $X$ has the following property: if $D$ is a dense subset of $X$, then for all open $U \subseteq X$, either $U$ is empty, or $|U \cap D| \geq \aleph_0$.

I'm a little confused about which spaces $X$ have this property. Clearly, if $X$ has an isolated point, then it can't have the property of interest. My question is whether the converse holds, and if not, what can we say?

Question. Is it true that if $X$ has no isolated points, then if $D$ is a dense subset of $X$, then for all open $U \subseteq X$, either $U$ is empty, or $|U \cap D| \geq \aleph_0$?
If not, what can we say?

 A: Assume that $X$ is Hausdorff, then the answer is positive. Suppose that $D$ is dense and $U$ is open, choose some $x,y\in U$ and find some $V,V'\subseteq U$ which are disjoint, with $x\in V$ and $y\in V'$.
Then by density $V\cap D$ and $V'\cap D$ are both non-empty. And therefore $U\cap D$ contains at least two points. By induction this shows that the intersection of $D$ and $U$ contains infinitely many points.
(I think that almost the same argument should also work for $T_1$ spaces, but I'm not entirely sure about $T_0$ spaces.)

If you don't require any separation, then taking $\Bbb R$ with the topology $\{(-n,n)\mid n\in\Bbb N\}\cup\{\Bbb R\}$ is a counterexample with $D=\{0\}$ as your dense set.
A: For a topological space $X$, the following are equivalent:


*

*$\{x\}$ is nowhere dense for every $x \in X$.

*Every finite subset of $X$ is nowhere dense.

*$D \cap U$ is infinite for every dense $D \subset X$ and every nonempty open $U \subset X$.
Proof:
$[1 \implies 2]$: A finite union of nowhere dense sets is still nowhere dense.
$[2 \implies 3]$: $D \cap U$ is dense in $U$. Therefore $D \cap U$ is infinite.
$[3 \implies 1]$: $\{x\} \cup (X\setminus \overline{\{x\}})$ is dense in $X$ and its intersection with $\operatorname{int} \overline{\{x\}}$ is a subset of $\{x\}$. Therefore
$\operatorname{int} \overline{\{x\}} =  \emptyset$. $\square$
In a T1 space, a singleton is nowhere dense unless it is open, so a T1
space without isolated points satifies 1-3.
