**ADDED LATER**

The answer to your question is that there is such a space $X$ if and only if $\mathfrak{b} = \aleph_1$.

*If $\mathfrak{b} = \aleph_1$, then there is such a space.*

To see this, first note that $\mathfrak{b} = \aleph_1$ if and only if there is an uncountable subset of the irrationals concentrated on $\mathbb Q$. This is proved by van Douwen in section 10 of his article in the *Handbook of Set Theoretic Topology*. By modifying his argument slightly, we can show that if $\mathfrak{b} = \aleph_1$ then there is an uncountable subset of the Cantor space concentrated on a countable subset of the Cantor space.

[The proof of this goes as follows. If $\mathfrak{b} = \aleph_1$, then one can construct via transfinite recursion a length-$\omega_1$ sequence $\langle f_\alpha :\, \alpha < \omega_1 \rangle$ of functions that is unbounded with respect to $\leq^*$, but also with the property that $\alpha < \beta$ implies $f_\alpha \leq^* f_\beta$. This means that a subset of this sequence is $\leq^*$-bounded if and only if it is countable. It's a well-known fact that the space $\omega^\omega$, endowed with the usual product topology, is homeomorphic to the space of irrational numbers. It is also fairly well known that the irrationals are homeomorphic to $C \setminus D$, where $C$ is the Cantor space and $D$ is some countable relatively dense subset of $C$. Let $Y$ be the image of your sequence of functions under some homeomorphism $\omega^\omega \rightarrow C \setminus D$. If $U$ is an open set containing $D$, then $C \setminus U$ is a compact subset of $C$. It's not too difficult to show that every compact subset of $\omega^\omega$ is $\leq^*$-bounded above by some function (or even more -- it is $\leq$-bounded), and this means $C \setminus U$ contains only countably many points of $Y$.]

Let $Y$ be, as in the previous paragraph, an $\aleph_1$-sized subset of the Cantor space $C$ that concentrates on a countable $D \subseteq C$. Let $X = Y \cup \mathbb Q$. As a subspace of the reals, this set $X$ meets all your requirements.

*A proof that if there is a space $X$ as described in your post then $\mathfrak{b} = \aleph_1$.*

Suppose there is an uncountable separable metrizable space $X$ that is concentrated on a countable set $D \subseteq X$. Let $\{ d_0, d_1, d_2, \dots \}$ be an enumeration of $D$. For each $x \in X \setminus D$, define a function $f_x : \omega \rightarrow \omega$ by setting $f_x(n) = \min\{ k :\, \mathrm{dist}(x,d_n) > \frac{1}{k} \}$.

Let $Y$ be a subset of $X \setminus D$ with $|Y| = \aleph_1$. The set of functions $\{ f_x :\, x \in Y \}$ is an $\aleph_1$-sized subset of $\omega^\omega$, and I claim it is unbounded with respect to $\leq^*$. In other words, I claim this set of functions witnesses $\mathfrak{b} = \aleph_1$.

To see this, suppose instead there is some $g \in \omega^\omega$ such that $f_x \leq^* g$ for all $x \in Y$. By a pigeonhole argument, there is some function $h \in \omega^\omega$, differing from $g$ in only finitely many places, such that $f_x \leq h$ for uncountably many $x \in Y$. Let $U = \bigcup_{n \in \omega} B_{1/h(n)}(d_n)$. This is an open set containing $D$, and our definition of the $f_x$'s ensures that $x \notin U$ if and only if $f_x \leq h$. Thus there are uncountable many $x$'s with $x \notin U$, contradicting the fact that $Y$ concentrates on $D$.

Let me point out that this argument is essentially contained in an earlier MO post by Taras Banakh found here. (A version of the argument seems to be in van Douwen's article too, and I don't know whether it really originated there either, but anyway I first learned it from Taras' post.)

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**ORIGINAL POST**

Here is a consistent example of a subspace of $\mathbb R$ with these properties.

First, there is an uncountable subset $X$ of $\mathbb R$ concentrating on $\mathbb Q$ if and only if $\mathfrak{b} = \aleph_1$. This is proved in chapter 10 of van Douwen's article in the *Handbook of Set-Theoretic Topology*. Note that $X$ concentrates on $\mathbb Q$ if and only if every subset of $X$ does, so if $\mathfrak{b} = \aleph_1$ then there is an $\aleph_1$-sized set of reals concentrating on $\mathbb Q$.

Now suppose $\aleph_1 = \mathfrak{b} < \mathrm{non}(\mathcal M)$ (the least size of a non-meager subset of $\mathbb R$). This situation is consistent -- it happens in the random real model, for example. In such a model, let $X$ be an $\aleph_1$-sized set concentrating on $\mathbb Q$. Adding countably many points to $X$ if necessary, we may (and do) assume $X$ is dense in $\mathbb R$. Then (viewed as a subspace of $\mathbb R$) it meets all your requirements.
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*A sketch of an argument as to why $\mathfrak{b} = \aleph_1$ implies there is an $\aleph_1$-sized set concentrating on $\mathbb Q$*:

Since you may not have access to van Douwen's article, here is a sketch of the idea I quoted above. If $\mathfrak{b} = \aleph_1$, then one can construct via transfinite recursion a length-$\omega_1$ sequence $\langle f_\alpha :\, \alpha < \omega_1 \rangle$ of functions that is unbounded with respect to $\leq^*$, but also has the property that $\alpha < \beta$ implies $f_\alpha \leq^* f_\beta$. It's a well-known fact that the space $\omega^\omega$, endowed with the usual product topology, is homeomorphic to the space of irrational numbers. Let $X$ be the image of your sequence of functions under any such homeomorphism. If $U$ is an open set containing $\mathbb Q$, then $\mathbb R \setminus U$ is a $\sigma$-compact subset of the irrationals. It's not too difficult to show that every $\sigma$-compact subset of $\omega^\omega$ is $\leq^*$-bounded above by some function, and this means $\mathbb R \setminus U$ can only contain countably many points of $X$.

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