Let $C$ be the Cantor Middle Third Set in $[0,1],$
Question Suppose that $\{O_1,O_2,...,O_n \}$ is a finite covering for $C,$ where each $O_i$ is an open interval for all $1\leq i \leq n.$
By rearrangement, we can assume that left endpoints of $O_i$ are in ascending order.
For any $1\leq i \leq n,$ does there exist $\alpha_i$ and $\beta_i$ such that $$[\alpha_i,\beta_i] \subseteq O_i, [\alpha_{i+1},\beta_{i+1}]\subseteq O_{i+1}, \alpha_i<\beta_i<\alpha_{i+1}<\beta_{i+1}, \bigcup_{I=1}^n [\alpha_i,\beta_i] \supseteq C$$ and $\alpha_i,\beta_i = \frac{m_i}{3^{n_i}}$ for some $m_i,n_i\in\mathbb{N}$ such that $m_i$ is not divisible by $3?$
I am trying to construct $\alpha_i$ and $\beta_i$ for $\{ O_1,O_2,O_3 \}.$
Let $O_1 = (c_1,d_1), O_2 = (c_2,d_2)$ and $O_3 = (c_3,d_3)$ be such that $c_1<c_2<c_3.$ Clearly $c_1<0.$ So we can choose $\alpha_1 = \frac{m_1}{3^{n_1}}$ for large enough $n_1$ so that $c_1<\alpha_1<0.$
Now, problem lies in getting $\beta_1.$ Clearly we need $\beta_1$ and $\alpha_2$ such that there is no element $x$ in Cantor set that lies between $\beta_1$ and $\alpha_2.$ Otherwise, $C$ may not be contained in the union of $[\alpha_i,\beta_i].$ But I have no idea how to choose such $\beta_1.$
I think this can be done as $C$ is nowhere dense, that is, for every interval $I,$ there is a subinterval $J\subseteq I$ such that $J\cap C = \emptyset.$
Put $\mathcal O=\{O_1,O_2,...,O_n \}$ and $O=\bigcup\mathcal O$. Since $\{C_m\}$ (see the Wikipedia reference) is a family of compact sets and $C=\bigcap C_m\subset O$, there exists $C_m\subset O$ (this simple fact even can be referenced as a corollary of Šura-Bura Lemma). Pick $k\ge m$ such that $1/3^{k}$ is smaller than Lebesgue number of the covering $\mathcal O$ of $C_m$. For any segment $S$ constituting $C_k$ let $i(S)$ be the smallest $i$ such that $S\subset O_i$. For each $i$ put $C(i)=\bigcup\{S:i(S)=i\}$, and if $C(i)\ne\varnothing$ put $\alpha_i=\min C(i)$, and $\beta_i=\max C(i)$. The choice of $k$ implies that $C_k\subset\bigcup C(i)$, the convexity of $O’s$ implies that $[\alpha_i,\beta_i]\subset O_i$ for each $i$, also the construction implies that $\beta_i<\alpha_{j}$ for each $i<j$. At last, it is easy to check that we can consecutively insert small segments $[\alpha_j,\beta_j]\subset O_j$ with ternary rational endpoints (necessarily disjoint with $C_k$) for these $j$ with empty $C_j$ before already constructed $[\alpha_i,\beta_i]$ and after the last $[\alpha_i,\beta_i]$ (for instance, in $n=7$, $C(2)$, $C(3)$, and $C(7)$ are empty then we insert $[\alpha_3,\beta_3]$ before $[\alpha_4,\beta_4]$, then $[\alpha_2,\beta_2]$ before $[\alpha_3,\beta_3]$, and then $[\alpha_7,\beta_7]$ after $[\alpha_4,\beta_4]$). Finally, we have $C\subset C_k\subset\bigcup[\alpha_i,\beta_i]\subset O$.