I am having a hard time solving this problem:
Let $f: [a, b] \rightarrow \Bbb R$ be a weakly increasing function, then: $$ \forall \epsilon > 0, \exists \delta > 0 : \forall y,x \in [a, b], y-x < \delta$$ implies that one of the following two conditions is met: $$f(y)-f(x) < \epsilon$$ $$ \exists z \in [x, y]: f(z_+)-f(z_-) \geq \epsilon, (f(y)-f(z_+)) + (f(z_-)-f(x)) < \epsilon $$ (where $f(z_-)$ and $f(z_+)$ are the left hand and right hand limits of $f$ at $z$ respectively). It isn't required that every pair at a distance less than $\delta$ satisfy just one condition, some of them may satisfy the first one, some others the second one.
So far I have been able to prove this for increasing functions with finitely many discontinuities (in which case I just re-defined the function to be continuous, hence uniformly continuous, by Heine-Cantor theorem, and then I used the $ \epsilon,\delta $ properties of the new function to find a suitable $\delta$) and for increasing functions with infinitely many discontinuities, provided that, the set of discontinuities has at most finitely many limit points.
The real problem arises if the increasing function has a set of discontinuities with infinitely many limit points, in which case my previous approach doesn't work. I think the solution has to do with the fact that I should exploit the hypothesis of monotonicity more, since I have just been using the existence of left and right hand limits at every point, which can also be satisfied by a non-monotone function.
Since I have already spent a lot of time thinking about this problem, I would really appreciate if someone could just push me toward the right direction without giving away too much, however, any help is obviously very welcome!
There is a typo in the question.
The function $f(x)=2x$ is a counterexample to the statement in your question. On any interval of length $\delta$, the value of $f(y)-f(x)$ will exceed $\delta$ somewhere. The possibility of a jump discontinuity is ruled out.
So, my guess is that the problem intended to say "$f(y)-f(x)< \epsilon$" and not "$f(y)-f(x)<\delta$".
$\\$
Now, for the problem as I think it was intended:
For each $x\in(a,b)$, either $f(x_+)-f(x_-) < \epsilon\;$ or $\;f(x_+)-f(x_-)\geq \epsilon$.
In the former case, you can find an open interval $(x-r,x+r)\subseteq[a,b]$ where $r_x>0$, such that $f(x+r)-f(x-r)<\epsilon$.
In the latter case, because $f(x_+)$ and $f(x_-)$ are limits, you can find an open interval $(x-r,x+r)\subseteq[a,b]$ where $r>0$, such that $(f(x+r)-f(x_+)) + (f(x_-)-f(x-r)) < \epsilon$.
As for $x=a$ or $x=b$, I'll leave you to find open intervals satisfying similar conditions and containing either of them. Recall that for $r>0$, the sets $[a,a+r),(b-r,b]$ are open subsets of $[a,b]$. If you don't know this, refer to https://en.wikipedia.org/wiki/Subspace_topology
Take all such intervals. Because we've accounted for every value $x\in[a,b]$, these open intervals cover $[a,b]$. Since $[a,b]$ is a compact set, a finite number of these open intervals are enough to cover $[a,b]$. Choose such a finite cover and call the intervals involved $I_1,I_2,\ldots,I_n$.
Because of the way these intervals will need to overlap, you can find $\delta>0$ such that for each closed interval of the form $[x,x+\delta]\subseteq[a,b]$, there exists $j$ such that $1\leq j\leq n$ and $[x,x+\delta]\subseteq I_j$. In other words, each closed subinterval of length $\delta$ in $[a,b]$ is contained in one interval $I_j$ where $j$ might of course depend on the subinterval. I'll let you think about how to prove this.
Furthermore, within each $I_j$, the conditions you want to prove are satisfied.