Is my proof of the Uniform Continuity Theorem good? I tried to prove the Uniform Continuity Theorem differently than my book does it, using completeness and the Cauchy Convergence Criteon rather than the Bolzano-Weierstrass theorem:

Theorem If a function $f$ is continuous on $[a, b]$, then it is uniformly continuous on $[a, b]$.
Proof.
Suppose that $f$ is not uniformly continuous on $[a, b]$. Then, the set $$S = \{ x \mid f \text{ is not uniformly continuous on } [a, x]\}$$ is non-empty since it contains at least $b$. Let $c = \inf S$. 
By the definition of the supremum, $f$ is not uniformly continuous on any neighborhood of $c$, meaning that there is an $\epsilon > 0$ such that any neighborhood of $c$ contains $x$ and $y$ for which $|f(x) - f(y)| > \epsilon$. By the Cauchy Convergence Criterion, $f$ must be divergent and therefore discontinuous at $c$.

Is my proof correct? What could I do to make it clearer and improve the style?
 A: This honestly makes very little sense to me. The sentence that begins “there must be an $\epsilon>0$” isn’t what you want to say at all if you’re using continuity. You want to say that for every $x,y,$ and $\epsilon$ there exists a $\delta$. However, even fixing the quantifier error I don’t see where the next sentence comes from and why it’s important. $|f(x)-f(y)|>0$ only tells you that $f(x)\neq f(y)$ and that’s going to hold on any interval on which $f$ is non-constant. It doesn’t tell you anything other than that $f$ is non-constant and I certainly don’t see why you suddenly conclude that $f$ diverges (I assume you mean here that there’s some point at which $f$ doesn’t exist and the left and right limits tend to infinity?)
You never use any property of $c$ or the set you define in this proof. You say that $x,y$ can be arbitrarily close to $c$ because $c$ is the sup but really $x$ and $y$ can be arbitrarily close to $c$ because the real numbers get arbitrarily close together.
Can you explain the thought process behind the proof in some detail?
A: I personally find your proof simple and clear, but you should start a new paragraph after "Let $c = \sup S$." and change:

There must be an $\epsilon > 0$ such that for any $\delta$ there are $x, y \in [a, b]$ for which $|x - y| < \delta $ and $|f(x) - f(y)| < \epsilon$. 
By the definition of the supremum, $x$ and $y$ can be made arbitrarily close to $c$, which means that any neighborhood of $c$ contains $x$ and $y$ for which $|f(x) - f(y)| > \epsilon$. By the Cauchy Convergence Criterion, $f$ must be divergent and therefore discontinuous at $c$.

to:

By the definition of the supremum, $f$ is not uniformly continuous on any neighborhood of $c$, meaning that any neighborhood of $c$ contains $x$ and $y$ for which $|f(x) - f(y)| > \epsilon$. By the Cauchy Convergence Criterion, $f$ must be divergent and therefore discontinuous at $c$.

This small change in wording makes everything much clearer in my opinion.
