If $f$ is monotone prove it is continuous 
Assume $f: [a,b] \to \mathbb{R}$ is a monotone function that satisfies the Intermediate Value Theorem. Prove that $f$ is continuous.

It is sort of confusing how they say it satisfies IVT. Don't only continuous functions satisfy IVT? If instead they mean the condition that if it takes any value between $[a,b]$ then it also take any value between $f(a)$ and $f(b)$, then we will need to use the definition of the monotonicity. Suppose that $x<y$ and $f(x) < f(y)$ for any $x,y \in [a,b]$.  How can we use this with IVT here to prove that $f$ is continuous? 
 A: It's not restrictive to assume that $f$ is increasing (otherwise use $-f$). Prove that, for $c\in(a,b)$,
$$
\lim_{x\to c^-}f(x)=\sup\{f(x):a\le x<c\},
\qquad
\lim_{x\to c^+}f(x)=\inf\{f(x):c<x\le b\},
$$
and that
$$
\lim_{x\to a^+}f(x)=\inf\{f(x):a<x\le b\},
\qquad
\lim_{x\to b^-}f(x)=\sup\{f(x):a\le x< b\}.
$$
Suppose that for some $c\in(a,b)$ the limits from the left and from the right are different and apply the hypothesis about the IVT. Finish up with the values at the extremes $a$ and $b$.
A: 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!
