Every continuous function $f: \mathbb{R} \to \mathbb{R}$ is uniformly continuous on every bounded set. Every continuous function $f: \mathbb{R} \to \mathbb{R}$ is uniformly continuous on every bounded set.
Here's what I did so far.  Let $U \subset \mathbb{R}$ be bounded set.  Let $\epsilon> 0$. For every $x \in U$, there is a $\delta_x$ such that if $|a - x| < \delta_x$, then $|f(x) - f(a)| < \epsilon$.  Now here's, where I'm not sure if my reasoning is correct.  We have
$$U \subseteq \bigcup_{x \in U} (x - \delta_x, x + \delta_x) $$
It seems intuitively clear that since $U$ is bounded, the union of only finitely many such intervals contains $U$, in which case we can simply let $\delta = \min\{\delta_x : x \in I\}$, where $I$ is a finite indexing set.  
Help please?
 A: There is an infinite number of items in your union, so you can't take the min so easily. You could have $\delta=0$. You have the good idea, but we need to construct the finite union. We will assume U closed, because we can always do the same on the closure of U. 
Given $\epsilon>0$ and $x_0=\min U$, we will define $\forall n,\ x_n=\max\left\{
{ x\in \left[x_{n-1}, \infty \right] } \mid
{ \lvert f(x)-f(x_{n-1})\rvert\leq\epsilon }
\right\}$.
Note that $x_n$ can be equal to $\infty$.
$\left(x_n\right)$ is stritly incresing, so converge to 
$x_\infty \in \mathbb R_+ \cup \{\infty\}$.
If 
$x_\infty \neq \infty$,
then $f$ is not continuous in $x_\infty$, which is not possible.
We have $U  $ bounded, and $x_n \rightarrow +\infty$, so $\exists n_0, \forall n>n_0, x_{n} \notin U $. Now $U \subset \bigcup_{i=0}^{n_0} [x_i, x_{i+1}]$ and you have your finite union.
A: Sorry to revive an old question , but I think there is a neat and elegant proof of it which also extends itself to a natural generalization : 
Let $f:\mathbb R \to \mathbb R$ be continuous and $A$ be bounded , then $\exists m>0$ such that $|x|<m , \forall x \in A$ , so $A \subseteq [-m,m]$ . Now $[-m,m]$ is compact in $\mathbb R$ , and  since $g:=f|_{[-m,m]}:[-m,m] \to \mathbb R$ is continuous so it is uniformly continuous , so $f|_A=g|_A:A \to \mathbb R$ is uniformly continuous . 
We note that if $A$ is a bounded subset of $\mathbb R^n$ then $\exists r>0 , a \in \mathbb R^n$ such that $A \subseteq B[a;r]$ , and since by Heine-Borel theorem , $B[a;r]$ is compact in $\mathbb R^n$ , so a similar restricting argument as above shows that : If $M$ is a metric space and $f:\mathbb R^n \to M$ is continuous , then $f$ is uniformly continuous on any bounded subset of $\mathbb R^n$ .
A: I believe this is covered by the "Heine-Cantor Theorem". The wikipedia page has an elegant proof based on the fact that every compact set has a finite subcover. 
https://en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem
