Delta in continuity Let $f: [a,b]\to\mathbb{R}$ be continuous, prove that it is uniform continuous.
I know using compactness it is almost one liner, but I want to prove it without using compactness. However, I can use the theorem that every continuous function achieves max and min on a closed bounded interval.

I propose proving that some choices of $\delta$ can be continuous on $[a,b]$, for example but not restricted to:
For an arbitrary $\epsilon>0$, for each $x\in[a,b]$ set $\Delta_x=\{0<\delta<b-a \;|\;|x-y|<\delta\Longrightarrow |f(x)-f(y)| <\epsilon\}$, denote $\delta_x = \sup \Delta_x $.
Basically $\delta_x$ is the radius of largest neighborhood of $x$ that will be mapped into a subset of neighborhood radius epsilon of $f(x)$. I'm trying to show that $\delta_x$ is continuous on $[a,b]$ with fixed $\epsilon$. My progress is that I can show $\delta_y$ is bounded below if $y$ is close enough to $x$, but failed to find its upper bound that is related to its distance with $x$.
Maybe either you could help me with this $\delta_x$ proof, or another cleaner proof without compactness (but allowed max and min). Thanks so much.

Let an $\epsilon>0$ be given and put
$$\rho(x):=\sup\bigl\{\delta\in\ ]0,1]\ \bigm|\ y, \>y'\in U_\delta(x)\ \Rightarrow\ |f(y')-f(y)|<\epsilon\bigr\}\ .$$
By continuity of $f$ the function $x\to\rho(x)$ is strictly positive and $\leq1$ on $[a,b]$.
Lemma. The function $\rho$ is $1$-Lipschitz continuous, i.e.,
$$|\rho(x')-\rho(x)|\leq |x'-x|\qquad \bigl(x,\ x'\in[a,b]\bigr)\ .$$
Proof. Assume the claim is wrong. Then there are two points $x_1$, $\>x_2\in[a,b]$ with
$$\rho(x_2)-\rho(x_1)>|x_2-x_1|\ .$$
It follows that there is a $\delta$ with $\rho(x_1)<\delta<\rho(x_2)-|x_2-x_1|$. By definition of $\rho(x_1)$ we can find two points $y$, $\> y'\in U_\delta(x_1)$ such that $|f(y')-f(y)|\geq\epsilon$. Now
$$|y-x_2|\leq |y -x_1|+|x_1-x_2|<\delta +|x_2-x_1| =:\delta'<\rho(x_2)\ .$$
Similarly $|y'-x_2|<\delta'$. It follows that $y$, $\>y'$ would contradict the definition of $\rho(x_2)$.$\qquad\quad\square$
The function $\rho$ therefore takes a positive minimum value $\rho_*$ on $[a,b]$. The number $\delta_*:={\rho_*\over2}>0$ is a universal $\delta$ for $f$ and the given $\epsilon$ on $[a,b]$.
