According to the book Advanced Calculus by Fitzpatrick:
The following example shows it is not true that a parametrized path $\gamma :[a, b] \to \mathbb{R^n}$ that has only the property that $\gamma :[a, b] \to \mathbb{R^n}$ is continuously differentiable is necessarily rectifiable: If the derivative is not bounded, the path may fail to be rectifiable.
Example 20.10 - First, define $$f(x) = \left\{ \begin{array}{ll} x \sin(\pi /2+\pi/x) & \mbox{if } 0 <x \leq 1 \\ 0 & \mbox{if } x = 0 \end{array} \right.$$ and then define $\gamma(t) = (t, f(t))$ for $0 \le t \le 1$. Then it is clear that $\gamma :[0, 1] \to \mathbb{R^2}$ is a parametrized path and that $\gamma :[0, 1] \to \mathbb{R^2}$ is continuously differentiable. However, the path $\gamma :[0, 1] \to \mathbb{R^2}$ is not rectifiable. Indeed, observe that $$f(1/k) = \left\{ \begin{array}{ll} 1/k & \mbox{if k is an even natural number} \\ -1/k & \mbox{if k is an odd natural number,} \end{array} \right.$$ so that $$||\gamma (1/k) - \gamma (1/(k+1))|| \ge (1/k) \text{for each natural number k.}$$ It follows that for each natural number $n$, if we define the partition $P= {\{0, 1/(n+1), 1/n, ..., 1/2, 1}\}$, then the polygonal approximation based on P is greater than $\sum_{k=1}^n 1/k$. Since the sequence of partial sums of the Harmonic Series $\sum_{k=1}^n 1/k$ is unbounded we conclude that the polygonal approximation can be arbitrarily large so the parametrized path $\gamma :[0, 1] \to \mathbb{R^2}$ is not rectifiable.
Question. By the same book we have the following
Definition. A parametrized path $\gamma :[a, b] \to \mathbb{R^n}$ is said to have arclength (or to be rectifiable) provided that there is a number $l$ with the following property: For each positive number $\epsilon$ there is a positive number $\delta$ such that $$|\sum_{k=1}^n|| \gamma(x_k) - \gamma(x_{k-1})||-l|< \epsilon$$ for each partition $P = {\{x_0, \dots, x_n}\} $ of$ [a, b]$ with $\text{gap} P < \delta$.
In the above definition of parametrized path to be rectifiable it is required that the $\text{gap} P < \delta$. However, in the example above gap is $1/2$ not a "small" number which normally depends on $\epsilon$ and becomes smaller as we choose $\epsilon$ to be small enough. Can the mentioned partition be eligible for the example above? Can the reason for the path to fail to be rectifiable be related to the wrong partition and not because of the unboundedness of the derivative of the parametrized path?
For a partition $P=\{a=x_0,x_1,\ldots,x_n=b\}$, let's write $$L_P(\gamma)=\sum_{k=1}^n\|\gamma(x_k)-\gamma(x_{k-1})\|.$$ We need the following observation: If a partition $P$ has a refinement $P'$, i.e. $P\subseteq P'$, then $L_P(\gamma)\le L_{P'}(\gamma)$, by triangle inequality.
It is then easy to see that if a path $\gamma$ is rectifiable, then $$\sup_P L_P(\gamma)=l<\infty.$$ (Please check it by yourself. Note here $P$ varies among all partitions of $[a,b]$.) I believe the converse is also true at least for $\gamma\in C^1$, but we don't need that.
Hence, to show that $\gamma$ is not rectifiable we only need to find a sequence of partition $P_j$ such that $L_{P_j}(\gamma)\to\infty$. So the argument in the example is valid.