Theorem : If a function $f$ is continuous on a closed and bounded interval $[a, b]$ then $f$ must be uniformly continuous in $[a, b]$
My Idea : I get the intuition that for a continuous function on a closed interval, range of $\delta$ for a given $\epsilon$ must be finite and in particular if we scan the function with 'window' of $\epsilon$ and obtain all deltas than $\inf \{\delta\}$ must be greater than zero.
Formalizing the idea
Let $\epsilon \in \mathbb {R} : \epsilon > 0$
Given, $f$ is continuous in $[a,b]$, therefore $\exists \delta (\alpha, \epsilon) > 0$, s.t.,
$|x-\alpha| < \delta \implies |f(x) - f(\alpha)| < \epsilon$
Now, denote $\delta (\epsilon) = \inf ~\{ \delta (\alpha, \epsilon) ~\forall ~\alpha \in [a,b] \}$
Now, the challenge remains to prove that this $\delta (\epsilon) > 0$ and we are (almost) done. But this is where I get stuck. I have an intuition that if delta goes zero than function must go asymptotic at that point but I am unable to formalize it.
I know there are other standrad proofs of this theorem, but I need to complete it this way because it is closer to what I imagine in my head.
Thanks
You can prove the uniform continuity theorem without invoking Heine-Borel. The only property related to compactness needed is that a continuous function on a closed, bounded interval attains its minimum at a point in the interval -- and this can be proved without any reference to open covers.
The reasoning is based on finding the infimum of a "maximal" delta for each $x \in [a,b]$ and so follows your intuition.
Suppose $f:[a,b] \rightarrow \mathbb{R}$ is continuous. Given $\epsilon > 0$, for each $x \in [a,b]$ let
$$\delta(x) = \sup \{ \delta>0: |f(y)-f(z)| < \epsilon \,\,\forall \,y,z \in [x-\delta/2,x+\delta/2]\}$$
Here $\delta(x)$ is the length of the largest interval $I(x)$ centered at $x$ such that $|f(y)-f(z)| < \epsilon$ when $y,z \in I(x)$.
We can dispense with the case $\delta(x) = \infty$ for some $x$ where any $\delta$ works for uniform continuity. If $\delta(x) < \infty$ for every $x$, we can show that $\delta(x)$ is continuous.
Consider the intervals centered at points, $x$ and $x + \omega$ where $\omega >0$. If $x+\omega - \delta(x+\omega)/2 \leq x -\delta(x)/2$ then $I(x) \subset I(x+\omega)$. In this case, the interval $I(x)$ could be made larger, which contradicts the construction of $I(x)$ as the maximal interval, and it follows that
$$x+\omega - \delta(x+\omega)/2 > x -\delta(x)/2 \implies \delta(x+\omega)-\delta(x) < 2\omega$$
Similarly, if $x+\delta(x)/2 \geq x +\omega +\delta(x+\omega)/2$ then $I(x+\omega) \subset I(x)$. Again, in this case, the interval $I(x+\omega)$ could be made larger, which contradicts the construction of $I(x+\omega)$ as the maximal interval , and it follows that
$$x+\delta(x)/2 < x +\omega +\delta(x+\omega)/2 \implies \delta(x+\omega)-\delta(x) > -2\omega.$$
Therefore $|\delta(x+\omega)-\delta(x)| < 2\omega$ for every $\omega >0$ and $\delta(x)$ is continuous.
Since $[a,b]$ is closed and bounded, a minimum value $\delta(c)$ is attained at some point $c$ in the interval.
Hence, for any $y,z \in [a,b]$, if $|y-z|< \delta(c)$ then $y,z \in [\xi-\delta(\xi)/2,\xi+\delta(\xi)/2]$ where $\xi = (y+z)/2$ and $|f(x)-f(y)|< \epsilon$. This proves that $f$ is uniformly continuous.