Intuitive way to understand the Smith–Volterra–Cantor set The Smith–Volterra–Cantor set (or ε-Cantor set) is a set of points on $ℝ$ that is nowhere dense, yet has positive measure.  
As I understand it, being nowhere dense means containing no intervals. An interval is a set of real numbers where any number that lies between two numbers in the set is also included in the set.  
How is it possible to have such a set on the real line where there are no intervals, yet the set still has positive measure? Can't we always simply pick two numbers from equal distance of any number in the set, thus creating an interval? Wouldn't removing all such options result in having no measure?
 A: "Can't we always simply pick two numbers from equal distance of any number in the set, thus creating an interval?" 
Why does that define an interval? Sure maybe that defines the endpoints of some interval--but the point is that the Smith-Volterra-Cantor set does not contain that interval
One definition for "no-where dense" is like you say:
$A\subseteq \mathbb{R}$ is a no-where dense set if and only if: 
For all intervals $J$ there exists a subinterval $I\subseteq J$ such that
$$
I \cap A = \emptyset 
$$
The point is that given any interval, we can find a subinterval inside of the given interval disjoint from $A$. Now, simply put, this means that given any interval $I$, $A\cap I$ is not dense in $I$. It does not mean that no-where denseness is equivalent to "containing no intervals". 
Take for example $\mathbb{Q}$. $\mathbb{Q}$ contains no intervals but is dense in every interval. ($\mathbb{Q}\cap I$ for any interval $I$ will also not contain intervals, but of course is dense in $I$). The reasons why these examples fail our definition is because there is no candidate subinterval that we can find because all intervals intersect $\mathbb{Q}$.
However, a no-where dense set in $\mathbb{R}$ does have the property of not containing an interval. 
Suppose there was an interval $K$ such that $A$ did contain $K$, as in, $K\subseteq A$. Well, by definition, there exists subinterval of $K$ that's disjoint from $A$--thereby showing that no such $K$ can exist. 
There's also an underlying topological notion here,  $A\subseteq \mathbb{R}$ is no-where dense, if and only if cl($A$) is no-where dense. This is easy to see, as if cl($A$) is not nowhere dense, then there exists an $I$ interval such that for all $J$ subintervals of $I$, $J\cap I \ne \emptyset$. In particular, every point of of $I$ must be contained in cl($A$).  As, for example if $I=[a,b]$ then, consider $x\in I$, let $J_N=[x-\frac{b-a}{kN},x+\frac{b-a}{kN}]\subseteq [a,b]$. Where $k$ is sufficiently large. Every interval $J_N$ intersects cl($A$) in something, hence, we can extract a cauchy sequence in $A$ converging to $x$ from these subintervals, thus $x\in cl(A)$ because $cl(A)$ is closed and contains its limit points. This proof can be adapted for the endpoints. Thus, we get that $cl(A)$ contains $I$, which means that $A$ was dense in $I$ to begin with. For the other direction, if $cl(A)$ is no-where dense, then $A$ is no-where dense. Let $I$ be an interval, then there exists a $J$ such that $J$ is disjoint from $cl(A)$, then $J$ is disjoint from $A$ as $A\subseteq cl(A)$. 
Your reasoning above is equating having positive measure with containing intervals. The whole point of the Smith-Volterra-Cantor set is that it's an explicit example that shows these two things are not the same.
"How is it possible to have such a set on the real line where there are no intervals, yet the set still has positive measure?"
Answer: Smith-Volterra-Cantor set.
A: If you remove all rational numbers, certainly what is left contains no intervals. If you take $\varepsilon>0$ and list all rationals $\{q_n:n\in\Bbb N=1,2,..\}$ and remove an interval centered at $q_n$ of length $\varepsilon/2^n$, then the total length (or measure) removed will be at most $\varepsilon$, so what is left has large measure. On the other hand, what is left is nowhere dense, since its complement (consisting of all open intervals centered at rationals that were removed) is open and dense. 
So if $C=[0,1]\setminus\cup_{n\in \Bbb N}\, (q_n -\frac\varepsilon{2^{n+1}},q_n +\frac\varepsilon{2^{n+1}})$, then $C$ is nowhere dense, but has measure $\ge 1-\varepsilon$. Specifically, for the Smith–Volterra–Cantor set one is careful in the construction to also ensure that $C$ is perfect (that is has no isolated points), but that is not difficult to achieve, in a similar way in which it is done for the usual Cantor set. 
