I'm taking a first course in differential geometry and want to make sure I really understand every definition. I know the modern definition(s) of manifolds arose (at least in part) from the desire to eliminate reference to embeddings in Euclidean space, but I'm going to try and go along with the course anyway.
A curve was not defined, but a parametrized curve was defined to be a continuous map $\gamma:[a,b]\to \mathbb R^3$. Then, the arclength of a parametrized curve was defined to be the supremum over all partitions of $[a,b]$ of polygonal approximations of the image of $\gamma$. If we assume $\gamma$ is $C^1$, some calculus leads to the convenient integral formula which coincides with a physicist's intuition and all is well.
However, the object of geometric interest here is really the image $A$ of $\gamma$, and the above is not "just" a polygonal approximation - we're taking a partition of the interval $[a,b]$ and then letting $\gamma$ pick the points on its image from which we construct the polygonal approximation. Given a drawing of a curve on a piece of paper one would simply pick some points on it and start connecting them - without any parametrization. This way one is led to define the approximate length of $A$ with respect to a choice of finite points $\mathcal C$ as $\ell(A,\mathcal C)=\sum_{x_i\in \mathcal C}\|x_i-x_{i-1}\|$ and the to define $$\ell (A)=\sup_\mathcal{C} \left\{ \ell (A,\mathcal C ) \right\}.$$ This looks much more obvious and intrinsic. It's also reminiscent of Lebesgue measure, except we're not approximating by covers but by polygonal lines.
The value of this supremum seems the identical to the classic definition, but I wonder, why isn't this one used? I think the reason is that parametrized curves basically give access to "physics" i.e we can integrate speed to get distance, but I just want to make sure.
The problem with your definition is that it doesn't work. :)
Consider the interval from $(0,0)$ to $(1,0)$ in the plane, and pick the sequence of points
$(1/3, 0)$, $(2/3, 0)$, $(1/4,0)$, $(3/4, 0)$, $(1/5, 0)$, $(4/5,0), \ldots ,(1/n, 0)$, $(1-1/n, 0)$.
The sum you propose will be at least $n/3$, so the upper bound on these sums, as $n \to \infty$, becomes infinite, while that segment surely has length 1.
Now the reason this goes wrong is that the points I've written down are "in the wrong order", but how can you conveniently specify that a sequence of points is in the "right" order, esp. for a curve that might look like the letter $\alpha$, i.e., might contain a crossing? The more you try, the more you realize that the thing you need is a parameterization.