Let $U$ be an open subset of $\mathbb{R}$, such that $f:U \rightarrow \mathbb{R}$ is a uniformly continuous homeomorphism. Show that $U=\mathbb{R}$. Here is my attempt at a proof :
Since $U$ and $\mathbb{R}$ are homeomorphic, $U$ must be connected, i.e an interval.
Case $1$ : U is bounded Let $a, b \in \mathbb{R}$ such that $U=(a,b)$. Then, we claim that $f$ can be extended continuously and uniquely to a continous function on $[a,b]$. Let ${\lbrace x_n \rbrace} \subset U$ converge to $a$, thus ${\lbrace x_n \rbrace}$ is Cauchy. By uniform continuity, ${\lbrace f(x_n) \rbrace}$ is Cauchy and hence convergent to, say, $l \in \mathbb{R}$, by the completeness of $\mathbb{R}$. Let ${\lbrace y_n \rbrace}$ be another sequence in U which converges to $a$ and let $\epsilon>0$ be given. By uniform continuity, $\exists \delta>0 \in \mathbb{R}$, sucht that $|x-y|<\delta \implies |f(x)-f(y)|<\epsilon.$
Then, since $\lim_{n\to\infty}(x_n-y_n)=0, \exists N \in \mathbb{N}$ such that, $|x_m-y_m|<\epsilon$ whenever $m>N$ and hence $|f(x_m)-f(y_m)|<\epsilon, m>N$. Thus, for any sequence ${\lbrace x_n \rbrace} \rightarrow a$, we have ${\lbrace f(x_n) \rbrace} \rightarrow l$. Similarly, we can show that for any sequence ${\lbrace z_n \rbrace}$converging to $b$, we must have ${\lbrace f(z_n) \rbrace} \rightarrow L$, for some $L$.
Define $$F:[a,b] \rightarrow \mathbb{R}$$ $$F(a)=l$$ $$F(x)=f(x), x\in U$$ $$F(b)=L$$
Then, $F$ is continuous on $[a,b]$ and a unique extension of $f$. Moreoever, $f(U) \subset F([a,b])$. Note however that $F([a,b])$ is a compact subset of $\mathbb{R}$ and hence not equal to it.
Case $2$: $U$ is an interval of the type $(a, \infty)$ or $(-\infty, b)$. In this case, an argument very similar to the one above shows that $f$ can be extended to a continuous function defined including the end-point. Thus, again, $f(U)$ will be a proper subset of $\mathbb{R}$.
Hence, $U=\mathbb{R}$.
Notation: $f[B]=\{f(b): b\in B\}$ when $B\subseteq dom (f).$
A homeomorphic image of a connected space is connected, and $f^{-1}:\Bbb R\to U$ is a homeomorphism so $U$ is a connected subspace of $\Bbb R,$ so if $U\ne \Bbb R$ then $U$ is a bounded open interval or an open half-line.
Suppose $U\ne \Bbb R.$
A uniformly continuous real image $f[U]$ of a bounded real interval $U$ must be a bounded subset of $\Bbb R$ (see $**$ below) so $U$ cannot be bounded.
Suppose $r\in \Bbb R$ and $U=(r,\infty).$ Consider the set $S=f[(r,r+2)],$ which is bounded, as $f$ is uniformly continuous. There exist $x,y\in [r,\infty)$ with $f(x)=-1+\inf S$ and $f(y)=+1+\sup S.$ Now $f$ is continuous and $f(x)<\inf S\le f(r+1)\le \sup S<f(y)$ so there exists $z$ between $x$ and $y$ with $f(z)=f(r+1).$ But $z>\min(x,y)>r+1$ so $f$ is not a bijection so $f$ is not a homeomorphism.(Contradiction.)
The case $U=(-\infty,r)$ is handled similarly.
$**.\,$ Appendix. If $U$ is a non-empty bounded real interval and $f:U\to \Bbb R$ is uniformly continuous, take $d>0$ such that $\forall a,b\in U\,(|a-b|\le d\implies |f(a)-f(b)|\le 1).$
Take $n\in \Bbb N$ with $nd\ge\sup U - \inf U.$ Choose some $x_0\in U.$ Now for any $x\in U$ there is an $m\in \Bbb N$ with $m\le n$ and an increasing or decreasing finite sequence $(x_0,...,x_m)$ of members of $U$ with $x_m=x$ and with $|x_{j+1}-x_j|\le d$ for $0\le j\le m-1.$ Then $|f(x)|=|f(x_m)|\le |f(x_0)|+\sum_{j=0}^{m-1}|f(x_{j+1}-f(x_j)|\le |f(x_0)|+md\le |f(x_0)|+nd.$