If $f$ is uniformly continuous on $(a, b)$, then $f$ is bounded on $(a, b)$.
I'm trying to understand this proof given here in this link. Its the second proof given from the question in the link, here I copy it for convenience:
Let $x_n\to1$, then $\{x_n\}_{n\in\mathbb{N}}$ is a Cauchy sequence, and so is $\{f(x_n)\}_{n\in\mathbb{N}}$, due to uniform continuity. This implies that $f$ extends continuously to $x=1$ and similarly to $x=0$. Hence, $f$ extends to a continuous function on a compact set $[0,1]$. Therefore, $f$ is bounded. ■
Can my sequences $X_n$ converge to a and then similarly another sequence converge to $b$ so that $f$ extends to a continuous function on the compact set $[a,b]$ thus $f$ is bounded on $(a,b)$? Why does the original user specifically state there sequence to converge to $0$ and another to $1$? Is my version of the proof acceptable?
Yes, your version is correct. The original user probably just wanted to prove for the specific case when $a=0$ and $b=1$. But that proof works for all $a,b\in\mathbb{R}$ when $a<b$.