If $x$ is a boundary of $S$ in $\mathbb{R}$, then $x$ must contain both interior points and exterior points of $S$ Above is the statement that I am given to prove or disprove.
I think it is false.
For $Q$ a rational number, there is no interior point nor exterior point. so every point in $Q$ is boundary point, but every ball of any point in $Q$ does not contain both interior and exterior points of $S$.
Is it valid counter-example to it? 
And I am wondering if $Q$ consists of $\textrm{int} S + \textrm{ext} S + \textrm{boundary} S$, where $S$ is subset of $Q$.
It is true for $R$, but not sure whether it still hold for $Q$.
 A: Be careful here. The boundary of $S=\mathbb{Q}\subset\mathbb{R}$ is actually $\partial S = \mathbb{R}$. You're right in saying that $\mathbb{Q}$ is part of the boundary, but if you drew an open ball around any irrational number, it would also contain both points in $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$. This is still a valid counterexample though, since — as you pointed out — $\mathbb{Q}$ has no exterior/interior points, so $\mathbb{R}$ still doesn't contain any.
A simpler counterexample might be the following set:
$$S = \{x\in\mathbb{R}: x<a\} = (-\infty, a).$$
You can check that $\partial S = \{a\}$, and that $a$ is not an interior point, so it also contradicts the statement.
A: If $S\subset\mathbb R$, then the boundary of $S$ is $\partial S=\overline A\cap\overline{A^c}$. So it is true that $\partial S$ contains limit points of $A$ and $A^c$. But unless $A$ and $A^c$ are both closed (i.e $A=\overline A$ and $A^c=\overline{A^c}$, it is not the case that $S\cap A\ne\varnothing$ and $S\cap A^c\ne\varnothing$. In fact, the only sets that satisfy this condition (being both open and closed) are $\varnothing$ and $\mathbb R$.
