Your idea is good. We already know that a continuous strictly increasing function is a homeomorphism. If we can find such a function $f:[0,1]\to [0,1]$ which carries the complement of one set onto the complement of another, this function will also carry one Cantor set onto the other.
But the task of piecing an infinite collection of linear functions together, checking the continuity of the whole thing, is rather tedious. I suggest building $f$ as a uniform limit of continuous maps $f_n$, following the process of creation of Cantor sets themselves. That is, let $C_n$ and $F_n$ be pre-Cantor sets of $n$th generation; with $C_0=F_0=[0,1]$. Let $f_0(x)=x$. Having defined $f_n$, proceed to $f_{n+1}=f_n$ on $[0,1]\setminus C_n$ and $f_{n+1}$ being an appropriate three-chain broken line on each component of $C_n$.
Alternatively, define $f_n$ as the unique increasing piecewise linear function on $[0,1]$ that maps the boundary points of $C_n$ bijectively onto the boundary points of $F_n$, and only changes the slope at these points.
This is essentially what you do already with linear pieces, with the difference that each $f_n$ is defined on all $[0,1]$.
By construction, $\sup |f_{n+1}-f_{n}| \le \max_k |J_{n,k}|\le 2^{-n}$. (I use $|\cdot|$ to denote length, and assume $J_{n,k}|$ is the same for all $k$, as on the picture.) Hence $\sup |f_{n+1}-f_{n }|\le 2^{-n}$, which implies that the sequence $f_n$ converges uniformly, by the Weierstrass $M$-test applied to $\sum (f_{n+1}-f_n)$. Let $f$ be its limit. Since
$f_n(C_m)\subset F_m$ for all $n\ge m$, we have $f(C_m)\subset F_m$. Hence $f(C)\subset F$.
Everything in the preceding paragraph applies to the inverse maps $f_{n}^{-1}$: namely, $\sup |f_{n+1}^{-1}-f_{n}^{-1}| \le \max_k |I_{n,k}|$, and so on. Let $g= \lim_{n\to\infty} f_n^{-1}$. Then both
$g\circ f = \lim_n g_n\circ f_n$ and $f\circ g = \lim_n f_n\circ g_n$ are identity maps. So, $g$ is the inverse of $f$.
This is not true: For every $p$, $\exists\ a,b$ such that $C \cap (a,b) = \{p\}$.
For example, $\frac 14 \in C$. In fact, there exists a non-constant sequence of endpoints in $C$ that converges to $\frac 14$. That sequence is: $$a_n = \sum_{k=1}^n \frac{2}{3^{2k}} \quad \implies \quad a_n \to \frac{1}{4}$$
This shows that $C$ does not have the discrete topology, because it has a limit point.
Best Answer
Yes, the Cantor set is totally disconnected. It does not follow (and it's not true) that every point is isolated. Not every function defined on the Cantor set is continuous.
You could use a Cantor-set-ish construction. This being just a hint, suppose $0\le f < 1$. Let $A_0$ be the set where $0\le f<1/2$ and $A_1$ the set where $1/2\le f<1$.
Choose compact sets $K_j\subset A_j$ of almost full measure. Now let $A_{0,0}=\{x\in K_0:0\le f(x)<1/4\}$ and $A_{0,1}=\{x\in K_0:1/4\le f<1/2\}$...