Rudin's proof that the Cantor set has no segments

We want to show that there is an integer $k$ such that $\alpha\le\frac{3k+1}{3^n}$ and $\frac{3k+2}{3^n}\le\beta$,

so we want to have $\frac{\alpha(3^n)-1}{3}\le k\le\frac{\beta(3^n)-2}{3}$.

Such an integer $k$ will exist if $\frac{\beta(3^n)-2}{3}-\frac{\alpha(3^n)-1}{3}\ge1$, so

$(\beta-\alpha)3^n\ge4$ gives $3^{-n}\le\frac{\beta-\alpha}{4}$.

Your denominators should be $3^n$, not $3^{-n}$.

Suppose that $\alpha$ is just a little bigger than $\frac{3k+1}{3^n}$ for some $k$, so that $(\alpha,\beta)$ is just barely too far to the right to contain $\left(\frac{3k+1}{3^n},\frac{3k+2}{3^n}\right)$ no matter how big $\beta-\alpha$ is. The next interval of that form to the right is

$$\left(\frac{3k+4}{3^n},\frac{3k+5}{3^n}\right)\;;\tag{1}$$

if we ensure that $\frac{3k+5}{3^n}\le\beta$, the interval $(\alpha,\beta)$ will contain the interval $(1)$. If $\alpha=\frac{3k+1}{3^n}+\epsilon$, then

$$\frac{3k+5}{3^n}=\alpha-\epsilon+\frac4{3^n}\;,$$

so we need to make sure that $\beta\ge\alpha-\epsilon+\frac4{3^n}$ no matter how small $\epsilon$ is. This will certainly be the case if $\beta\ge\alpha+\frac4{3^n}$, i.e., if $\beta-\alpha\ge\frac4{3^n}$. In other words, if we choose $n$ big enough so that

$$\frac1{3^n}\le\frac{\beta-\alpha}4\;,$$

the interval $(\alpha,\beta)$ is certain to contain an interval of the form $\left(\frac{3k+1}{3^n},\frac{3k+2}{3^n}\right)$.

Clearly this will be the case if $\frac1{3^n}\le\frac{\beta-\alpha}6$.