How to show that the natural logarithm is Lipschitz on $[\beta, \infty)$
Since we are dealing with the logarithm, we can use its properties. Assume without loss of generality that $\beta \leq x \leq y$: $$ |\ln x - \ln y| = \ln\frac{y}{x} = \ln\!\left(1+\left(\frac{y}{x}-1\right)\right) \leq \frac{y}{x}-1 = \frac{1}{x}\left(y-x\right) \leq \frac{1}{\beta}\left(y-x\right) = \frac{1}{\beta}|y-x| $$ where we only used the inequality $\ln(1+u)\leq u$ for all $u > -1$. (Which is standard, and can be proven e.g. by concavity of $\ln$).
The derivative, being $1/x$ is upper bounded by $1/\beta$ and also monotone decreasing on this domain.