Every interval has a subinterval contained in $A'$ iff $A'$ contains a dense open set iff $\overline{A}$ has no interior point Assume that all sets below are subsets of $\mathbb{R}.$
In page $2$ of Oxtoby's Measure and Category, he stated the definition of nowhere dense set. 
$A\subseteq\mathbb{R}$ is nowhere dense if it is dense in no interval, that is, if every interval has a subinteraval contained in the complement of $A.$
Subsequently, he quoted the following: 

$A$ is nowhere dense if and only if its complement $A'$ contains a dense open set, and if and only if its closure $\overline{A}$ has no interior points. 
Question: How to prove the three equivalences, that is, 
$(1)$ Every interval has a subinterval contained in the complement of
  $A,$
$(2)$ The complement of $A$ contains a dense open set, 
$(3)$ Closure $\overline{A}$ of $A$ has no interior points.

I want to show that $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1).$
I have no idea on how to start any of the implication. Any idea would be appreciated. 
 A: $(1)\Rightarrow (2)$ : Let us prove that the interior of the complement of $A$ is dense. Let $I\subset \Bbb R$ be any non-empty open interval; then $I$ has a subinterval contained in the complement of $A$. WLOG we can assume this subinterval to be open, and then it must also be contained in the interior of the complement of $A$. Thus $I\cap (\Bbb R\setminus A)^{o}\neq \emptyset$.
$(2)\Rightarrow (3)$ : If $\Bbb R\setminus A$ contains a dense open set, then its interior must be dense. Thus
$$\Bbb R= \overline{(\Bbb R\setminus A)^{o}}=\overline{\Bbb R\setminus \overline{A}}=\Bbb R \setminus (\overline{A})^{o},$$so $(\overline{A})^{o}$ is empty, i.e. $\overline{A}$ has no interior point.
$(3)\Rightarrow (1)$ : Assume for a contradiction that there is some interval $I$ with no subinterval contained in the complement of $A$, that is, every subinterval of $I$ intersects $A$. Then $I\subset \overline{A}$, which contradicts the hypothesis that $\overline{A}$ has no interior point.
A: We'll show that $\neg(2) \implies \neg(1)$. This is equivalent to $1 \implies 2$.
Suppose the complement of $A$ does not contain a dense open set. Consider $A^\circ$, the interior of $A$. This is the largest open set contained in $A$. Since this is not dense, there is a point $p$ that does not lie arbitrarily close to it i.e. $\exists \epsilon >0 , \forall x \in A^\circ, |x - p| > \epsilon$. Then, clearly the interval $(p-\frac\epsilon 2, p+\frac \epsilon 2)$ is not intersecting with $A^c$, since it keeps strictly positive distance from $A^\circ$. Hence, such an interval is entirely contained in $A$.
We will show that $2 \implies 3$
If the complement of $A$ has a dense open set, and $p$ is an interior point of $A$, then there is a neighbourhood of $p$, call it $V$, such that $V \subset A$. But then, $V$ does not intersect the complement of $A$, and this contradicts the fact that the complement has a dense open set.
Next, $3 \implies 1$.
So the closure of $A$ has no interior points. Let $(x,y)$ be an interval. If this interval has no subinterval contained in the complement of $A$, then it is entirely contained in $A$. However, this means any point in $(x,y)$ is an  interior points of $A$ and hence of the closure of $A$, which is a contradiction. Hence, $(x,y)$ must have a subinterval contained in  the complement of $A$. 
A: 2 and 3 are equivalent whatever the space.
3 is interior closure A is empty which is equivalent to
closure interior complement A is whole space (#2) simply
by taking complements and useing the important theorem
complement interior K = closure complement K.  
