Is this proof correct? If so, then is my presentation good enough? If not, then where lie the flaws?
The presentation is easy to read and understand. I have corrected the flaws (see my edit of your question). Now the proof is correct, but extremely long, because it consists of a big number of short steps. I don’t have read Rudin, because I learned analysis by old Russian and Ukrainian books, and I don’t know what level of mathematical rigor is required for it, but such detalization level is more like to logical profs than to mathematical that. :-) So you may try to work in formal logic, maybe you’ll like that. :-)
Is the formulation of this result correct and general enough? If so, then is my proof (and the presentation thereof) good enough?
Yes, all is OK.
Is this proof correct? If so, then what is the presentation like?
The proof is correct, the presentation is easy to read and understand.
Is every real convex function $f$ defined in $(a,b)$ also uniformly continuous?
Not necessarily. For instance, a function $f(x)=1/x$ defined on $(0,1)$ is convex, but not uniformly continuous.
Let $x<y$ be arbitrary points in $[c,d] \subset (a,b)$. Take $\delta >0$ such that $a+\delta < c < d < b-\delta$.
If $f$ is convex on $(a,b)$, then it is easy to show that it is bounded on any closed subinterval. Hence, there exist bounds $m$ and $M$ such that $m \leqslant f(x) \leqslant M$ for all $x\in [a+\delta,b- \delta].$
Take a fixed $z$ such that $d < z \leqslant b-\delta$ and define $\lambda = \frac{y-x}{z-x}$. It follows that $0 < \lambda < 1$ and $y = \lambda z + (1-\lambda)x$, and by convexity
$$f(y) \leqslant \lambda f(z) + (1-\lambda)f(x) = f(x) + \lambda(f(z) - f(x))$$
Hence,
$$f(y) - f(x) \leqslant \lambda(f(z) - f(x)) \leqslant \lambda (M - m) = \frac{y-x}{z-x} (M-m) < \frac{y-x}{z-d} (M-m) < \frac{M-m}{z-d}|y-x|$$
Switching variable names $x$ and $y$ we get
$$-[f(y) - f(x)] = f(x) - f(y) \leqslant \frac{M-m}{z-d}|x- y| = \frac{M-m}{z-d}|y- x|,$$
and this implies
$$|f(y) - f(x)| \leqslant \frac{M-m}{z-d} |y-x|.$$
Therefore, $f$ is continuous on $(a,b)$ as well as Lipschitz continuous on any closed subinterval.
Best Answer
The pictorial version. (But it is the same as your inequality version, actually.)
Suppose you want to prove continuity at $a$. Choose points $b,c$ on either side. (This fails at an endpoint, in fact the result itself fails at an endpoint.)
By convexity, the $c$ point is above the $a,b$ line, as shown:
Again, the $b$ point is above the $a,c$ line, as shown:
The graph lies inside the red region,
so obviously we have continuity at $a$.