Rudin's PMA, theorem 4.20. A continuous function which is not uniformly continuous on a set which is not closed

Relevant portion from Rudin's PMA To make explicit the argument that $f$ is not uniformly continuous, I tried the following. I wanted to ask if it looked alright.

Consider $$f(x) = \frac{1}{x - x_0}$$ Here, $x_0$ is a limit point of $E$ and $x_0 \notin E$. We claim that is not uniformly continuous on $E$. Take $\epsilon = 1$. We will show that given $\delta > 0$, there exists $t_\delta, x_\delta \in E$ with $|t_\delta - x_\delta|< \delta$ such that $|f(x_\delta) - f(t_\delta)| \geq 1$.

Choose $x_\delta$ such that $|x_\delta - x_0| < \delta$. Note that the $\delta$-neighbourhood of $x_\delta$ is a neighbourhood of $x_0$, so it contains points arbitrarily close to $x_0$. Thus there exists $t_\delta$ with $|x_\delta - t_\delta| < \delta$ such that $|t_\delta - x_0| < \frac{1}{1 + k}$ where $k = \frac{1}{|x_\delta - x_0|}$. Then $$|f(x_\delta) - f(t_\delta)| \geq |f(t_\delta)| - |f(x_\delta)| = \left|\frac{1}{x_0 - t_\delta}\right| - \left|\frac{1}{x_0 - x_\delta}\right| \geq 1$$

Your proof is fine. However, I would not have used the letter $k$ in this context, for psychological (not logical) reasons. It's just that $k$ usually represents an integer, but here it can be any positive real.