I prefer to add picture in my question since this is not my proof. I was able to follow the proof but I lost the marked line when it said the cluster set of $a_n$ from the left is $[-1,1].$ This is equivalent to say that $\{(a_n,x)\colon x\in[-1,1]\}$ is a subset of the limit points of the graph of $f.$ Notice that $f(0)=0$ and $f(1)=0.$ Can we say also $(0,0)$ and $(1,0)$ are limit points of the of $f$? let's just take $a_n=\frac{1}{3}$ and $b_n=\frac{2}{3}.$ since $f$ is continuous so the graph of $f$ is closed. Then, $(\frac{1}{3},-1)$ is limit point of the graph. My understanding for the marked line is that $\{(\frac{1}{3},x)\colon x\in[-1,1]\}$ is the limit point of the graph but this is not correct of course.
Any help will be appreciated greatly. 
For your case $a_n = \frac{1}{3}$, $b_n = \frac{2}{3}$, to understand the "cluster set of $a_n$ from the left" you need to consider a sequence of components of $I \setminus C$ which converge to $a_n$ from the left, namely: $$\left(\frac{1}{9},\frac{2}{9}\right), \left(\frac{7}{27},\frac{8}{27}\right), \left(\frac{25}{81},\frac{26}{81}\right),\left(\frac{79}{243},\frac{80}{243}\right),\ldots $$ The portion of the graph over each of these intervals is a line segment whose $y$-coordinate varies from $-1$ at the left endpoint to $+1$ at the right endpoint. If you sketch those line segments you will see that they are getting steeper and steeper, with slopes approaching $\infty$; the $\Delta x$ of each line segment is approaching zero, whereas the $\Delta y$ of each line segment is approaching $1 - (-1)=2$. Also, those line segments are accumulating on the segment $$\frac{1}{3} \times [-1,+1] = \{(1/3,x) \mid -1 \le x \le +1\} $$ which you observed in your post. Therefore, for each $x \in [-1,+1]$ you can choose a sequence of points $$p_1 \in \left(\frac{1}{9},\frac{2}{9}\right), \,\, p_2 \in \left(\frac{7}{27},\frac{8}{27}\right), \,\,p_3 \in \left(\frac{25}{81},\frac{26}{81}\right), \,\,p_4 \in \left(\frac{79}{243},\frac{80}{243}\right), ... $$ such that $\lim_{n \to \infty} f(p_n) = x$. In other words, "the cluster set of $a_n$ from the left is $[-1,+1]$".
Added: To see a bit more about how to choose the sequence $p_n$, I'll consider separate cases.
If $x \in (-1,+1)$ then $p_1 \in \left(\frac{1}{9},\frac{2}{9}\right)$ is uniquely determined so that $f(p_1)=x$, and similarly for $p_2,p_3,p_4$ and so on; thus, $f(p_n)=x$ for all $n$ and therefore $\lim_{n \to \infty} f(p_n)=x$.
If $x = +1$ then we have to choose $p_n \in (a_n,b_n)$ more carefully: choose $$p_n = \frac{1}{n} a_n + \frac{n-1}{n} b_n $$ Working through the algebra you will get $$f(p_n) = \frac{1}{n}(-1) + \frac{n-1}{n}(1) = 1 - \frac{2}{n} $$ and therefore $\lim_{n \to \infty} f(p_n)=1$.
If $x=-1$ then you can figure out a similar way to choose $p_n \in (a_n,b_n)$ so that $\lim_{n \to \infty} f(p_n)=-1$.