The orignal question is to prove that-"If $f$ has Intermediate Value Property and is monotonic then $f$ is continous"
I was looking for a proof and here is a proof by fellow mathematician.
An $\epsilon,\delta$-proof:
Let $x_0 \in (a,b)$ be arbitrary and let $\epsilon > 0$. Let $s_1 = \min(f(x_0)+\epsilon/2, f(b))$. Then the number $f(x_0)+s_1$ satisfies $f(x_0) < f(x_0)+s_1 < f(x_0)+\epsilon$. By the intermediate value property, there exists then some $\delta_1 > 0$ such that $f(x_0 + \delta_1) = f(x_0)+s_1$ (by the choice of $s_1 = \min(f(x_0)+\epsilon/2, f(b))$ we are guaranteed to stay within the domain of $f$). Similarly, let $s_2 = \max(f(x_0)-\epsilon/2, f(a))$ and find a $\delta_2>0$ such that $f(x_0-\delta_2) = f(x_0)-s_2$. Take $\delta = \min(\delta_1,\delta_2) > 0$.
Now consider the punctured neighbourhood $U_\delta = \{x \in (a,b) : 0 < |x - x_0| < \delta\}$. Without loss of generality, we may assume that $f$ is increasing. Let $x \in U_\delta$ be arbitrary. If $x > x_0$ we have, as $f$ is increasing, $f(x_0) \leq f(x) \leq f(x_0+\delta_1) = f(x_0)+s_1 < f(x_0)+\epsilon$, whence $|f(x)-f(x_0)| < \epsilon$. If $x < x_0$, we have, as $f$ is increasing, $f(x_0)-\epsilon < f(x_0)-s_2 = f(x_0-\delta_2) \leq f(x) \leq f(x_0)$, whence $|f(x)-f(x_0)| < \epsilon$.
You can work out the proof for the endpoints as an exercise!
I cannot understand why $f(x_0)+s_1$ satisfies $f(x_0) < f(x_0)+s_1 < f(x_0)+\epsilon$, If this is correct then $s_1<\epsilon$
if $f(x_0)+\epsilon/2 < f(b)$ then $s_1=f(x_0)+\epsilon/2 < \epsilon$ Hence $f(x_0)<\epsilon$
What is wrong here.
The first paragraph as written is incorrect. Take $f(x)=x$ on $[0,1]$, $x_0=\frac 12$ and $\epsilon \in (0,1/2)$. Then $f(x_0)+ s_1 = 1/2+1/2 + \epsilon/2 > f(1)$.
Here is another approach. By monotonicity, the one-sided limits exist at all points, therefore only discontinuity is a jump discontinuity. A jump discontinuity contradicts the intermediate property.