Arc length of the Cantor function

Possibly the easiest way to see this is to note that any partition divides $[0,1]$ into two kinds of intervals: those on which $f$ is constant, and those on which it is not. The total length of the constant intervals can be made arbitrarily close to 1, while the total length of the nonconstant intervals in any partition is at least 1 because they add up to a displacement of 1 on the $y$-axis. The result follows.

Partitions of the form $\{\frac{k}{3^n}:0\leq k\leq n\}$ would do the job of showing that the arclength is at least $2$. I don't know what you did exactly, but notice that the partition $\{0,1\}$ shows that the arclength is at least $\sqrt 2$. The partition $\{0,\frac{1}{3},\frac{2}{3},1\}$ shows that the arclength is at least $\frac{1}{3}(\sqrt{13}+1)$. And so on; but we would need some good way to keep track of things to see that the sum goes to $2$ as $n$ goes to infinity.

Edit: The previous version of my answer had a serious error, which has been fixed.

Here is an approach using the symmetry of the function. Let $A$ denote the desired arclength, and for each positive integer $k$, let $A_k$ denote the arclength of the restriction to $\left[0,\frac{1}{3^k}\right]$. I assert without proof something that is geometrically clear from the graph: $A_k=2A_{k+1}+\frac{1}{3^{k+1}}$. This comes from splitting the interval $\left[0,\frac{1}{3^k}\right]$ into thirds, and noticing that the portions of the graph on the outside thirds are congruent, while on the middle third the graph is a horizontal segment. This leads to $A$ being expressed in terms of $A_k$ as $$A=2^kA_k+\frac{1}{3}\sum_{j=0}^{k-1}\left(\frac{2}{3}\right)^j,$$ which is easily proved by induction using the prior equation. Notice that $A_k\geq \frac{1}{2^k}$ for all $k$, and $\displaystyle{\sum_{j=0}^{k-1}\left(\frac{2}{3}\right)^j}$ goes to $3$ as $k$ goes to infinity, so letting $k$ go to infinity shows that the arclength is at least $2$.