Is it true that any subset of countable points chosen from a connected metric space has no interior points? Let us take the connected metric space $(\mathbb{R},d)$ where $d(x,y)$ may be defined as $d(x,y)=|x-y|$. 
Now let us define the subset $\mathbb{N}$ of $\mathbb{R}$. No matter how we choose $r$, every neighborhood $N_r(x)$ where $x$ is a natural number, would contain all real numbers in $(x-r,x+r)$, and hence would not be entirely contained in $\mathbb{N}$. Hence no point in the subset $\mathbb{N}$ of $\mathbb{R}$ can be an interior point. 
By this logic it seems to me that if we define a subset of countable points chosen from a connected metric space, it would not have any interior points. 
Is that correct?
 A: Suppose $A$ is a countable subset of a connected metric space $M$, and $x\in M$; we'll show $x$ is not an interior point of $A$.
Towards contradiction, suppose $x\in int(A)$. Then there is some positive $\epsilon$ such that $B_\epsilon(x)\subseteq A$. But since $A$ is countable, and the interval $(0, \epsilon)$ is uncountable, there must be some $\delta<\epsilon$ such that $d(x, a)\not=\delta$ for any $a\in A$. 
Since $B_\epsilon(x)\subseteq A$ and $\delta<\epsilon$, this means in fact that $d(x, y)\not=\delta$ for any $y\in M$.
But now we can write $M=I\sqcup O$, where 


*

*$I=\{y\in M: d(x, y)<\delta\}$, and

*$O=\{y\in M: d(x, y)>\delta\}$.
Each of $I$ and $O$ is open, so this contradicts the connectedness of $M$.
A: As posted, the answer is "NO." Here is an example. Take
 $ X =  \mathbb{R}\times \{0\} \cup \mathbb{N}\times \{1\}  $
i.e.: $X$ is the x-axis in $\mathbb{R}^2$, union $\mathbb{N}$ at height 1. Clearly $X$ is uncountable, and with the usual distance in $\mathbb{R}^2$, all points on the countable subset $\mathbb{N}\times \{1\}  $ are interir points, since for any $ n \in \mathbb{N}  $ , the ball centered at $p=(n,1)$ of radious 0.5, which reduces to the set $\{p\}$,  is contained in $\mathbb{N}\times \{1\}  .$
