Counterexamples abound. Here are a few of them.

**Theorem.** If $\emptyset\ne X\subseteq\mathbb R$ and $X\subseteq\operatorname{cl}(\operatorname{int}(X))$, then $\mathcal D(X)\cong\mathcal D(\mathbb R)$.

**Proof.** Construct an infinite sequence of pairwise disjoint open intervals $I_n$ so that $\bigcup_{n=1}^\infty I_n$ is a dense subset of $X$, and $|X\setminus\bigcup_{n=1}^\infty I_n|=2^{\aleph_0}$. Then a set $D\subseteq X$ is dense in $X$ if and only if $D\cap I_n$ is dense in $I_n$ for each $n$.

Thus, if $X'$ is another set satisfying the same hyotheses, with an analogous sequence of intervals $I'_n$, then a bijection $f:X\to X'$ which maps each $I_n$ homeomorphically onto the corresponding $I'_n$ is an isomorphism from $\mathcal D(X)$ to $\mathcal D(X')$.

**P.S.** With a somewhat more complicated argument one can prove:

**Theorem.** If $X$ and $Y$ are nonempty Polish spaces with no isolated points, then $\mathcal D(X)\cong\mathcal D(Y)$.