Any subspace of a separable metric space is separable and hence admits a countable dense subset. $\mathbb R$ is separable, since $\mathbb Q$ is a countable dense subspace.
Thus, if $A\subseteq\mathbb R$, we can find a countable dense subset $D$ of $A$. We clearly don't have $\overline{A\cap\mathbb Q}=A$ (just take $A=\left\{\sqrt 2\right\}$, for example), but what if $A=[0,\infty)$ or $A$ is any other connected set (i.e. an interval)?
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The answer to the question in the title is "yes". Quick proof: take the decimal expansion of any $0 \le x< \infty$, and consider approximating this decimal expansion by its first digit, its first two digits, etc. This is a sequence in $\Bbb Q$ that converges to $x$. Hence $x$ is in the closure of $\Bbb Q$.
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If A is any open set in $\mathbb{R}$, for every point $a \in A$ you have an interval $(a - \epsilon,a + \epsilon) \subseteq A$. Using density of the rationals in $\mathbb{R}$, just pick a sequence of rationals that converges to $a$. Since it converges, just keep the "tail" of it, i.e. pick a subsequence that falls in $(a - \epsilon,a + \epsilon)$. Hence $A\cap \mathbb{Q}$ is dense in $A$. The same can be done if $A$ is a closed interval.
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First of all,if $S$ is a enumerable and conected subset of real line it must be a singleton.
Fix your favorite connected subset of real line $S$ that is not a single point (this case is ruled out because its dense in itself)
If you agree that $A=\mathbb{Q}\cap[0,1]$ is dense on $[0,1]$ and that $(A+1)=(a+1 : a\in A)$ is dense in $[1,2]$ then we may agree that $A_z=(A+z)$ is dense in $[z,z+1]$ for $z \in \mathbb{Z}$.
Now taking $Q = \cup_{z\in\mathbb{Z}}(A_z\cap S)\subset\mathbb{Q}$ can you tell me if it's a suitable dense subset of $S$?
In general, if $(\Omega,\tau)$ is a topological space, $Q\subseteq\Omega$ is $\tau$-dense and $A\in\tau$, then $A\cap Q$ is $\left.\tau\right|_A$-dense. The simple proof can be found here.