You ask: "Is Wikipedia's definition of fractal the standard?" and right near the top of Wikipedia's page of fractals, we see the following definition:
A fractal is a mathematical set that has a fractal dimension that
usually exceeds its topological dimension and may fall between the
integers.
The statement that the fractal dimension may "fall between the integers" really adds nothing but, other than that, I would say that this is fairly standard; it is unquestionably the definition that was put forward by Mandelbrot around 1975 when he coined the term "fractal". He did not refer to "fractal dimension" at that time but, rather, the "Hausdorff-Besicovitch dimension" as he put it. In fairness, the usefulness of this definition has been debated with even Mandelbrot himself feeling that it might not be inclusive enough. Nonetheless, this comparison of dimension is central in fractal geometry. Gerald Edgar calls his great book, Measure, Topology, and Fractal Geometry, a meditation on the definition.
Taking this to be the definition, we can definitely say that the Cantor set satisfies it. If by "fractal dimension" you mean similarity dimension, then the Cantor set has fractal dimension $\log(2)/\log(3)$, since it's composed of two copies of itself scaled by the factor three. Also, the set is regular enough that any reasonable definition of fractal dimension agrees with that computation. (Well, any real-valued defintion.)
Topological dimension is a trickier thing, actually. It's inductive in nature. Totally disconnected sets (like single points, finite sets, or notably the Cantor set) have dimension zero. Higher dimensions are defined in terms of lower dimensions. The space we live in is three dimensions because balls in this space have a surface that is two dimensional. Because of this inductive nature, topological dimension always yields an integer.
When you write that you "do not see the irregular aspects or the complexity that is usually inherent with fractals", I think you might have a bit of a mis-understanding about fractal geometry. The Cantor set is indeed regular but, then so are all the strictly self-similar sets studied in classical fractal geometry - the Koch curve, the Sierpinski triangle, the Menger sponge, and countless others all display this regularity. Indeed, it's exactly this regularity that allows us to understand them.
To emphasize this regularity, and how it appears in not just the Cantor set, compare the following zooms of
The Cantor set
The Koch curve
Now, of course, there are "irregular" fractals - or, at least, less regular fractals. Examples include random version of self-similar sets, examples that arise from number theory, and examples arising from complex dynamics (like Julia sets). It's not their irregularity that makes these objects fractal, however. On the contrary, its the regularity that we can find that allows us to analyse these objects to the point where we can characterize them as fractal. Of course, this analysis is bit harder with these less regular examples.
"The elements in the Cantor set are the end points of all the intervals in $E_n$..." This is your mistake. This isn't true. In fact, written in ternary expansion, the elements of the Cantor set are precisely those elements in $[0,1]$ with a ternary expansion consisting of $0$'s and $2$'s (where we note $0.01=0.00\bar{2}\in\mathcal{C}$, but $0.0101\notin\mathcal{C}$, for example). Using this fact, it isn't hard to show that $\frac{1}{4}\in\mathcal{C}$ but $1/4$ is not an endpoint of any interval.
Best Answer
Convergence in fractal geometry is typically defined in terms of the Hausdorff metric. Roughly, two sets are "close" with respect to the Hausdorff metric, if every point in one is close to some point of the other.
Your collection of Cantor sets is indeed dense in the Koch curve with respect to the Hausdorff metric. The Hausdorff metric, however, doesn't respect length. That is, two sets can be close in the Hausdorff metric, yet their lengths can be very far apart. You've found one example illustrating this but there are others.
For example, if $$ Q_n = \{k/n:0\leq k \leq n\}, $$ then the Hausdorff distance between $Q_n$ and the unit interval is less than $1/n$. $Q_n$ is finite, yet the sequence of $Q_n$s converges to a set of positive length. Similarly, you could strengthen your example by using the set of endpoints of the intervals that approximate the Koch curve.
Here is a strategy to find points in the Koch curve that do not lie on any of your Cantor sets. First, note that the Koch curve is invariant set of the iterated function system:
$$\begin{align} T_1(x,y) &= \left(\frac{x}{3},\frac{y}{3}\right) \\ T_2(x,y) &= \left(\frac{1}{6} \left(x-\sqrt{3} y+2\right),\frac{1}{6} \left(\sqrt{3} x+y\right)\right) \\ T_3(x,y) &= \left(\frac{1}{6} \left(x+\sqrt{3} y+3\right),\frac{1}{6} \left(-\sqrt{3} x+y+\sqrt{3}\right)\right) \\ T_4(x,y) &= \left( \frac{x}{3},\frac{y+2}{3} \right). \end{align}$$
These functions map the Koch curve onto the four sub-parts below
Now, any point On the Koch curve can be realized as the limit of a sequence $$\begin{align} & T_{i_1}(0,0) \\ & T_{i_1} \circ T_{i_2}(0,0)\\ & \vdots \\ & T_{i_1} \circ T_{i_2} \circ \cdots T_{i_n}(0,0) \\ & \vdots \end{align}$$ where $(i_1,i_2,i_3,\ldots)$ is a sequence in $\{1,2,3,4\}$. The point lies on one of your Cantor sets precisely when the sequence contains only finitely many 2s and 3s, so that it ends in a string of only 1s and 4s.
If, for example, the sequence has only 1s and 4s, and no 2s or 3s, then we get a point in Cantor's ternary set lying on the unit interval. If the sequence start with a 2 and then contains only 1s and 4s, we generate a point in the red Cantor set shown below; this is exactly the image of the ternary Cantor set under the function $T_2$. If the sequence starts 3, then 2, then contains only 1s and 4s, we generate a point in the blue Cantor set below; this is exactly the the image of the ternary Cantor set under the function $T_3 \circ T_2$.
Finally, if we have any other sequence of 1s, 2s, 3s, and 4s, then we generate some other point on the Koch curve that is not in any of your Cantor sets. There are uncountably many such points. I suppose the simplest one to find explicitly corresponds to the sequence containing only 2s, which is exactly the fixed point of $T_2$. To find it, we need only solve $$T_2(x,y) = (x,y),$$ which yields $(5/14,\sqrt{3}/14)$. That point is shown in red in the sequence of approximations below.
If we zoom into the last picture on the red point so that it's centered in a square of side length 0.04, we get the following:
Thus, the edges keep jutting out closer to the point but never actually hit it. It's in the limit but not on any of the edges.