After defining the Cantor set $P$, Rudin claims no segment of the form $$\Big(\frac{3k+1}{3^m},\frac{3k+2}{3^m}\Big)\tag{24}$$ where $k$ and $m$ are positive integers, has a point in common with $P.$ (it took me a while to verify this; I found this argument and accepted despite wanting more!)
Then he claims : since every segment $(\alpha,\beta)$ contains a segment of the form $(24)\dots$
I wanted to verify this i.e $\Big(\frac{3k_o+1}{3^{m_o}},\frac{3k_o+2}{3^{m_o}}\Big) \subseteq (\alpha,\beta) \subseteq (0,1) $ for some positive integers $k_o$ and $m_o$.
Attempt: For the possibility of inclusion we would want the length of $\Big(\frac{3k+1}{3^m},\frac{3k+2}{3^m}\Big)$ which is $\frac{1}{3^m}$ to be less than the length of $(\alpha,\beta)$ which is $\beta -\alpha$. By the Archimedean property it is possible to find an $m_o(\in \mathbb{N}) > \log_3\frac{1}{\beta - \alpha}$ so that $\frac{1}{3^{m_o}}<\beta -\alpha.$
But I have trouble finding the right $k_o.$ It is possible again to choose $k_o$ by Archimedean property to have $\frac{3k_o+1}{3^{m_o}}> \alpha.$ But how do you ensure that $\frac{3k_o+2}{3^{m_o}}< \beta?$
As you observed the length of the interval $\big(\frac{3k+1}{3^m},\frac{3k+2}{3^m}\big)$ is $\frac{1}{3^m}$. Now given $\alpha<\beta\in\mathbb{R}$, let $m$ be sufficiently large so that $\frac{1}{3^m}<\frac{\beta-\alpha}{4}$. This choice of $m$ implies that $$\text{($\star$) }\beta-\frac{4}{3^m}>\alpha$$
From now on $m$ is fixed and we have to choose $k$ (which is going to be dependent on $m$). Since $\frac{3k+1}{3^m}\rightarrow\infty $ as $k\rightarrow\infty$ there exists $k$ for which $\frac{3k+1}{3^m}>\alpha$, let $k_0$ be the minimal $k$ satisfying this condition.
It is left to show that the interval $\big(\frac{3k_0+1}{3^m},\frac{3k_0+2}{3^m}\big)$ lies in $(\alpha,\beta)$. We already know that $\alpha<\frac{3k_0+1}{3^m}$ hence it left to show that $\frac{3k_0+2}{3^m}<\beta$.
Suppose not, then $\frac{3k_0+2}{3^m}\geq\beta$, hence $\frac{3k_0-2}{3^m}+\frac{4}{3^m}\geq \beta$ we conclude that $\frac{3k_0}{3^m}\geq\beta-\frac{4}{3^m}$ From ($\star$) we have that $\frac{3(k_0-1)+1}{3^m}>\alpha$ which is a contradiction to the minimality of $k_0$. This completes the proof.