How can complement of the Cantor set be the countable union of open intervals?

The cardinal arithmetic does not work like that. The fact that each set is of cardinality bigger than $2^n$ is not really important, and certainly does not translate to $2^{\aleph_0}$ at infinify.

We still have a finite set at each step. And countable union of finite sets is countable. In your particular case infinite countable, i.e. $\aleph_0$. And that is regardless of their concrete size.

Not yet explicitly pointed out, the error you made was in assuming that $\sup_{i∈ℕ} 2^{c_i} = 2^{\sup_{i∈ℕ} c_i}$ for every sequence of cardinals $(c_i)_{i∈ℕ}$. Be careful not to swap operations unless they really can be swapped, such as supremum and the real exponentiation function. Note on the other hand that $\sup_{i∈ℕ} 2^{k_i} = 2^{\sup_{i∈ℕ} k_i}$ for every strictly increasing sequence of ordinals $(k_i)_{i∈ℕ}$. Ordinal exponentiation is different from cardinal exponentiation; in particular $ω = 2^ω$.

You may be overthinking this with the set cardinality equation. You've described an enumeration of the open intervals. In fact, since each interval has a distinct left-hand endpoint that is a rational number, this gives an injection from the set of intervals into the rationals, showing that the set is countable.

But we can say more if we like. After the first step, there's $1$, after the second step, there are $2$ more for a total of $1 + 2 = 3$, and after the $n$th step, there are $2^n - 1$ of them. This is a bijection from the naturals to the set of open intervals.

If we want to make it explicit, we have to write down a breadth-first enumeration of the infinite complete binary tree. Each node of the tree at depth $n-1$ corresponds to one of the $2^{n-1}$ new intervals of size $3^{-n}$ that we remove at step $n$. (Sketch a picture of, say, step $n = 3$ with $2^2 = 4$ new intervals, each of width $\frac{1}{27}$ if this is difficult to visualize.)

The bijection takes a positive integer $k$ and produce an interval $I_k = (a_k, b_k)$. Using the observation that node/interval $I_k$ will have two children $I_{2k}$ and $I_{2k+1}$, it seems natural to represent $k$ in binary: $$ k = \sum_{j=0}^{n-1} b_j \, 2^j, $$ where $2^{n-1} \leq k < 2^n$. The most significant bit $b_{n-1} = 1$ is treated separately. The endpoints are easy to describe: \begin{align} a_k &= \frac{1}{3^n}\biggl(1 + 2\sum_{j=1}^{n-1} b_{j-1} \, 3^j \biggr) \\ b_k &= \frac{1}{3^n}\biggl(2 + 2\sum_{j=1}^{n-1} b_{j-1} \, 3^j \biggr) \end{align}

So, for example, with $n = 3$, the indices are $k$ such that $4 \leq k < 8$ and the intervals are: \begin{array}{cccc} k & b_2b_1b_0 & a_k & b_k \\ \hline 4 & 100 & \frac{1}{27} \bigl( 1 + 2(0 + 0) \bigr) = \frac{1}{27} & \frac{1}{27} \bigl( 2 + 2(0 + 0) \bigr) = \frac{2}{27} \\ 5 & 101 & \frac{1}{27} \bigl( 1 + 2(0 + 3) \bigr) = \frac{7}{27} & \frac{1}{27} \bigl( 2 + 2(0 + 3) \bigr) = \frac{8}{27} \\ 6 & 110 & \frac{1}{27} \bigl( 1 + 2(9 + 0) \bigr) = \frac{19}{27} & \frac{1}{27} \bigl( 2 + 2(9 + 0) \bigr) = \frac{20}{27} \\ 7 & 111 & \frac{1}{27} \bigl( 1 + 2(9 + 3) \bigr) = \frac{25}{27} & \frac{1}{27} \bigl( 2 + 2(9 + 3) \bigr) = \frac{26}{27} \end{array}