How to prove that if $f$ is Lipschitz of order $\alpha>0$ on an interval, then $f$ is uniformly continuous on this interval? Spivak's Calculus, chapter 11 "Significance of the Derivative", problem 37:



*A function $f$ is Lipschitz of order $\alpha$ at $x$ if there is a constant $C$ such that

$$(*)\ \ \ \ \ |f(x)-f(y)| \leq C|x-y|^{\alpha}$$
for all $y$ in an interval around $x$. The function $f$ is Lipschitz
of order $\alpha$ on an interval if $(*)$ holds for all $x$ and $y$
in the interval.
a) If $f$ is Lipschitz of order $\alpha>0$ at $x$ then $f$ is
continuous at $x$.
b) If $f$ is Lipschitz of order $\alpha>0$ on an interval, then $f$ is
uniformly continuous on this interval.

We can prove $a)$ as follows

Let $y=x+h$. Then $$|f(x)-f(x+h)| \leq C|h|^{\alpha}$$
$\forall \epsilon>0$ if $|h|< \left ( \frac{\epsilon}{C} \right
 )^{\frac{1}{\alpha}}$ then $|f(x)-f(x+h)|<\epsilon$. Therefore
$$\lim\limits_{h \to 0} f(x+h)=f(x)$$ Ie, $f$ is continuous at $x$.

My question is about proving $b)$.
The straightforward proof is

$\forall \epsilon>0$ choose $\delta=\left ( \frac{\epsilon}{C} \right 
> )^{\frac{1}{\alpha}}$. Then for all $x$ and $y$ in an interval, we
have $|f(x)-f(y)|<\epsilon$. Therefore $f$ is uniformly continuous on
the interval.
This is essentially taking the given interval and using it in the
definition of $f$ being Lipschitz of order $\alpha$ at each point in
the given interval.

Is it possible to have a slightly simpler proof using the following theorem (Spivak, Appendix to Chapter 8, "Uniform Continuity")?

Theorem 1: $f$ continuous on $[a,b] \implies f$ uniformly continuous on $[a,b]$

Here is a slightly incomplete alternative proof for b):

$f$ being Lipschitz of order $\alpha$ on an entire interval means it
is Lipschitz of order $\alpha$ on each point in the interval.
By a), $f$ is continuous at each point in the interval.
By Theorem 1 above, if the interval is closed then $f$ is uniformly
continuous on the interval.

The question is, can I make this latter proof complete by extending it to all intervals, not just closed intervals? What reasoning do I need to use?
Given an open interval, I don't think I can just take a closed interval enclosing it because I don't know if $f$ is defined in the enclosing interval.
 A: Yes, I do believe you can complete the proof in the way you are suggesting. However, I think it might require some heavy machinery.
Essentially, what you want is a Lipschitz extension, so that if $f:(a,b) \to \mathbb{R}$ is Lipschitz, you can extend it to $\hat{f}:[a,b] \to \mathbb{R}$ and keep it Lipschitz. Then you will know that $\hat{f}$ is uniformly continuous on $[a,b]$, and hence $f$ is uniformly continuous on $(a,b)$.
The classical Lipschitz extension theorem is Kirszbraun's Theorem, which works in the very general setting of Hilbert spaces, of which $\mathbb{R}$ is an example. This theorem promises that you can extend $f$ to all of $\mathbb{R}$ without changing the Lipschitz constant. Kirszbraun's theorem doesn't actually apply to $\alpha$-Lipschitz (sometimes called Hölder continuous) functions, but that is covered in Theorem 1 of this paper.
Now, these are deep results applying to vast generalizations of this problem, and I'm sure that there are simpler proofs in this setting. Still, if you wanted to extend $f$ to an enclosing interval, it is possible!
