Could someone verify my proof that if $f:(a,b) \rightarrow \mathbb{R}$ is uniformly continuous, then $f$ is bounded? For my proof, I had to inverse the definitions for uniform continuity and sequence convergence, so it feels a bit sketchy to me. I'd also appreciate suggestions if you have any.
Let $f:(a,b) \rightarrow \mathbb{R}$ be uniformly continuous. For the sake of contradiction, assume $f$ is unbounded. Let the sequences $a_n \rightarrow a$, $b_n \rightarrow b$, and $x_n \rightarrow x_0 \in (a,b)$. Since $f$ is continuous, then $f(x_n) \rightarrow f(x_0) \in \mathbb{R}$. Since $f$ is unbounded, then either of the sequence $f(a_n),f(b_n)$ must diverge, otherwise $f$ would be bounded. Without loss of generality, assume the sequence $b_n$ diverges. Since $f$ is uniformly continuous, then for $\epsilon > 0$, there exists $\delta > 0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$ for $x,y \in (a,b)$. Let $\delta > 0$. Since $b_n \rightarrow b$, then there exists some $N \in \mathbb{N}$ where $|b-b_n|<\delta$ for $n \geq N$. Restrict $x \in (a,b)$ such that $b_n<x<b$. Then $|x-b_n|<\delta$. Since $f(b_n)$ diverges, then there exists some $\epsilon > 0$ and $n \in \mathbb{N}$ with $n \geq N$ such that $|f(x)-f(b_n)| \geq \epsilon$. Therefore we can write for $|x - b_n|<\delta$ and $|f(x)-f(b_n)| \geq \epsilon$. Therefore $f$ is not uniformly continuous. This is a contradiction since $f$ is uniformly continuous. Therefore $f$ is bounded.

Suppose $f$ were unbounded on $(a,b).$ Let $x_1$ be any member of $(a,b).$ For $n\in \Bbb N$ let $x_{n+1}\in (a,b)$ with $|f(x_{n+1})|\ge 1+|f(x_n)|.$
Observe that this implies that $|f(x_m)-f(x_n)|\ge 1$ when $m\ne n.$
Let $(j(n))_{n\in \Bbb N}$ be some sub-sequence of $\Bbb N$ such that $(x_{j(n)})_{n\in \Bbb N}$ converges to some $x\in [a,b].$
Note that $m\ne n$ implies that $j(m)\ne j(n),$ which implies $|f(x_{j(m)})-f(x_{j(n)})|\ge 1.$
Then there does NOT exist $\delta>0$ such that $\forall u,v\in (a,b)\;(|u-v|<\delta \implies |f(u)-f(v)|<1).$
Because for any $\delta>0 $ the interval $I_{\delta}=(a,b)\cap (x-\delta/2,x+\delta/2)$ contains $x_{j(n)}$ for all but finitely many $n,$ so there exist $m,n$ with $m\ne n$ and $\{x_{j(m)},x_{j(n)}\}\subset I_{\delta}.$ And for such $m,n$ we have  $$|x_{j(m)}-x_{j(n)}|<\delta \;\text { but }\; |f(x_{j(m)})-f(x_{j(n)})|\ge 1.$$
