Let $U \subset \mathbb{R}$ ,open and $h:U \to \mathbb{R}$ , where h is a uniformly continuous homeomorphism.Then $U=\mathbb{R}$ 
Let $U \subset \mathbb{R}$ ,open and $h:U \to \mathbb{R}$ , where h is a uniformly continuous homeomorphism.Then $U=\mathbb{R}$

My attempt :
Let $p \in \mathbb{R}$ and $p$ not in $U$ be a limit point of $U$.I will try to show that $p \in U$,so that I can conclude that $U$ is both open and closed.
Now,  we create a sequence $\{u_n\} \to p$.Since it is a convergent sequence so it is a cauchy sequence. Since $h$ is uniformly continuous and $\mathbb{R}$ is complete so $\{h(u_n)\} \to h(p)$.Now,as it is a homeomorphism so it is an onto mapping hence $p \in U$.So $U$ is both open and closed.
Now, we will try to show that the only open and closed subset of $\mathbb{R}$ are $\phi$ and $\mathbb{R}$.Let $A$ be an open and closed subset of $\mathbb{R}$ then, as $A$ is open it can be written as the union of countable disjoint  segments in $\mathbb{R}$.
So,  $A=(a_1,a_2) \cup (a_2,a_3) ..\cup (a_{n-1},a_{n})$.So $a_2 \in A^c$ which is also a closed and open set of $\mathbb{R}$.Hence, $a_2$ is an interior point of the set $A^c$, which is not possible.
I don't think it is correct and I am missing something. Can someone just go through my proof.

First part requires  a small correction. You can only say that the Cauchy sequence $h(p_n)$ converge to some real number $x$.  Since $h$ is onto you can write $x$ as $h(q)$ for some $q\in U$. But continuity of the inverse map shows $p_n \to q$ which forces $p=q \in U$.
The fact that there are no open and close subsets of $\mathbb R$ other than  $\mathbb R$ itself and  the empty set is just connectedness of the real line.
