"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.
Best Answer
In the Smith-Volterra-Cantor set, the lengths of the intervals that get removed are multiplied by $1/4$ in each step. But, that's not what $a_j$ represents. Instead, $a_j$ represents the fraction of each of the remaining intervals that gets removed in the $j$th step.
So in the first step of the Smith-Volterra-Cantor set, you remove the middle $1/4$, and $a_1=1/4$. But in the next step, you remove intervals of length $1/4^2$ from each of the two remaining intervals. Those remaining intervals each have length $3/8$, so the fraction we're removing is $\frac{1/4^2}{3/8}=\frac{1}{6}$. That's $a_2$.
So the values of $a_j$ are changing and aren't always $1/4$. It turns out that they are shrinking fast enough that the product $\prod_{1}^\infty (1-a_j)$ ends up converging to a value greater than $0$, and so in the end $C$ has positive measure.