If $f$ is strictly increasing on $[a,b]$ then $f$ is continuous at some point in the interval. Edit: Proof version 2.0
Thanks to all for your insight and suggestions.
Theorem: If a function $f$ is strictly increasing on the closed interval $[a,b]$, then $f$ is continuous at some point in the interval.
Proof:
By assumption, $f$ is increasing on $[a,b]$. Note that $f$ is therefore bound on $[a,b]$; that is, for all $x$ in $(a,b)$ we have
$$f(a) < f(x) < f(b).$$
Furthermore, for all $x < y$ in $[a,b]$
$$0 < f(y) - f(x) \leq f(b) - f(a).$$
Lemma: For any $\varepsilon > 0$ and subinterval of $[a,b]$ (including $[a,b]$ itself), there are points $a'$ and $b'$ in the subinterval with $a' < b'$ and
$$f(b') - f(a') < \varepsilon.$$
Informally, the lemma says that there are no "minimum distances" between $f$ values over any intervals. Within any interval, we can always find distinct points with $f$ values that are arbitrarily close to each other.
Lemma Proof:
The lemma say there are no intervals with "minimum distances". Let us assume the opposite.
That is, we assume instead that there exists some number $\varepsilon'> 0$, and points $c$ and $d$ such that
$$a \leq c < d \leq b,$$
and for all $a'< b'$ in $[c,d]$,
$$f(b') - f(a') \geq \varepsilon'.$$
Now, there are an infinite number of points between $c$ and $d$, and this combined with the above suggests $f$ is unbound on $[a,b]$.
To put it more formally, first, we know that
$$f(a) \leq f(c) < f(d) \leq f(b).$$
For any $n \in \mathbb{N}$, we can divide the interval $[c,d]$ into a partition of $n$ parts: $P = \{t_0, t_1, t_2,\dots, t_n\}$, with $c = t_0 < t_1 < t_2 < \dots < t_n = d$.
We see then that
$$f(d) \geq f(c) + n\cdot \varepsilon',$$
$$f(b) \geq f(a) + n\cdot \varepsilon'.$$
Since $\varepsilon' > 0$ and $n$ can be chosen to be arbitrarily large, $f(b)$ (or $f(a)$) must be unbounded, which is absurd.
There must not be a subinterval with a "minimum distance" $\varepsilon'$. That is, for any subinterval of $[a,b]$ and any $\varepsilon > 0$, we can find $a'<b'$ in the subinterval with
$$f(b') - f(a') < \varepsilon.$$
$\blacksquare$
We return to the theorem proof. We begin by applying the lemma to the number $\varepsilon_0 = 1$ and the interval $[a,b]$.
The lemma shows that there exist numbers $a'$, $b'$ in $[a,b]$ such that $a \leq a' < b' \leq b$ with
$$f(b') - f(a') < 1.$$
Furthermore, for any two points $x < y$ in $(a', b')$, we have
$$f(y) - f(x) < 1.$$
We pick two such points and label them $a_0$ and $b_0$.
Thus, for any two points $x < y$ in $[a_0, b_0]$ we have
$$f(y) - f(x) < 1.$$
We can repeat this process, this time applying the lemma to $\varepsilon_1 = 1/2$ and the interval $[a_0, b_0]$.
This results in points $a_1$ and $b_1$ with
$$a < a_0 < a_1 < b_1 < b_0 < b,$$
and for all $x < y$ in $[a_1, b_1]$,
$$f(y) - f(x) < 1/2.$$
Continuing in this way $n$ times, we arrive at points
$$a < a_0 < a_1 <\dots<a_n < b_n < \dots < b_0 < b,$$
and for all $x < y$ in $[a_n, b_n],$
$$f(y) - f(x) < \frac{1}{2^n}.$$
Now, given any $\varepsilon > 0$, we can find some $n \in \mathbb{N}$ such that $\frac{1}{2^{n}} < \varepsilon$.
If we then do the above process $n+1$ times, we generate points
$$a < a_0 <\dots<a_{n+1} < b_{n + 1} < \dots < b.$$
The nested interval theorem tells us there is at least one point $c$ that sits within all intervals generated this way. More specifically,
$$a_n<a_{n+1} \leq c \leq b_{n + 1} < b_n,$$
or, $c$ is an interior point in $(a_n, b_n)$.
If we let
$$\delta_{n} = \min(c - a_{n},  b_{n} - c),$$
We see that for all $x$, if
$$|x - c| < \delta_n \text{ then } |f(x) - f(c)| < \frac{1}{2^n} < \varepsilon.$$
(By restricting $x$ to be sufficiently close to $c$ we guarantee $x$ sits in $(a_n, b_n)$, which of course also includes $c$.)
To reiterate, the point $c$ is in all such nested intervals.
For any $\varepsilon > 0$ we can find a $\delta > 0$ such that for all $x$ if
$$|x - c| < \delta \text{ then } |f(x) - f(c)| < \varepsilon.$$
Hence, $f$ is continuous at $c$.
Finally, note that we can apply the preceding arguments to any closed subinterval of $[a,b].$ If $f$ is increasing on $[a, b]$, $f$ is continuous at some point within any subinterval of $[a,b]$. The points at which $f$ is continuous form a "dense set".
In the case of decreasing $f$, we can apply the theorem to $g = -f$.
$\blacksquare$
As an aside, I just realized that this theorem comes freely without any extra work, from the combined results of two of the problems in Chapter 13 of Spivak's Calculus (3rd Edition).
Problem 13-20 shows that if $f$ is nondecreasing on $[a,b]$, $f$ is integrable on $[a,b]$.
Problem 13-30 shows that if $f$ is integrable on $[a,b]$, then $f$ must be continuous on a dense set on $[a,b]$.
Ah well. Nothing like the (long) road less travelled!
Original post
What follows is an ostensible proof that any strictly increasing function $f$ on a closed interval $[a,b]$ must be continuous at some point in $[a,b]$.
Note that if true, this can be generalized to decreasing functions, and to subintervals of $[a,b]$.
If correct, the general implication seems to be that any strictly increasing or strictly decreasing function on an interval is continuous on a dense set within the interval.
My background: I'm a total novice, just working by way (slowly) through Spivak's Calculus.
This proof is not from the book, or at least not from anything I've seen yet.
I stumbled upon it while thinking about another problem.
I thought I'd post it here to see if I'm making any obvious errors.
Theorem: If a function $f$ is strictly increasing on the closed interval $[a,b]$, then $f$ is continuous at some point in the interval.
Proof:
By assumption, $f$ is increasing on $[a,b]$. Note that $f$ is therefore bound on $[a,b]$; that is, for all $x$ in $(a,b)$ we have
$$f(a) < f(x) < f(b).$$
Lemma:  If $f$ is strictly increasing on $[a,b]$ and $f$ is discontinuous at every point on $[a,b]$, then there exists some $\varepsilon'> 0$ such that for all $x$, $y$ in $[a,b]$ with $x<y$,
$$f(y) - f(x) \geq \varepsilon'$$.
Informally, this says that there's some "minimum distance" between the $f$ values of distinct points. They aren't allowed to get arbitrarily close to each other. Basically, there's nothing but jump discontinuities!
Lemma Proof
Suppose instead that there is no "minimum distance", i.e. for any $\varepsilon_1 > 0$ there exist some $a_1$, $b_1$ in $[a,b]$ with $a_1 < b_1$ and
$f(b_1) - f(a_1) < \varepsilon_1.$
For this $\varepsilon_1$, $a_1$, $b_1$, we have for all $x$ in $(a_1, b_1)$,
$$f(a_1) < f(x) < f(b_1)$$.
Take the midpoint
$$m_1 = \frac{a_1 + b_1}{2},$$
and the number
$$\delta_1 = \frac{b_1 - a_1}{2} > 0.$$
We see for all $x$, if
$$|x - m_1| < \delta_1 \text{, then } |f(x) - f(m_1)| < \varepsilon_1.$$
Now, by assumption there's no "minimum distance", so we can apply the above arguments to any $\varepsilon > 0$.
For each choice $\varepsilon_i$, there is an $a_i$ and a $b_i$ in $[a,b]$, a corresponding midpoint $m_i$, and another $\delta_i>0$ such that for all $x$, if
$$|x - m_i| < \delta_i \text{, then } |f(x) - f(m_i)| < \varepsilon_i.$$
Since we can always find these $m_i$ midpoints and $\delta_i$'s, this implies the existence of some point $m'$ that works for every $\varepsilon > 0$.
(Note: I think this is shakiest part of my proof. Maybe there's a more airtight way of putting it, or maybe I'm wrong!)
Thus there exists some point $m'$ in $[a,b]$ (or in $(a,b)$, really) such that for any $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $x$ if
$$|x - m'| < \delta \text{, then } |f(x) - f(m')| < \varepsilon.$$
In other words, $f$ is continuous at $m'$, which contradicts the hypothesis that $f$ is discontinuous everywhere on $[a,b]$.
Our assumption that there is no "minimum distance" must be wrong if $f$ is both increasing and completely discontinuous on $[a,b]$.
If $f$ is increasing and discontinuous, then there must be a "minimum distance" $\varepsilon'$ between $f$ values.
$\blacksquare$
With the lemma in hand, we return to the general proof.
The lemma tells us if $f$ is increasing and discontinuous on all of $[a,b]$ then there exists some $\varepsilon' > 0$ such that for all $x < y$ in $[a,b]$,
$$f(y) \geq f(x) + \varepsilon'.$$
Now, there are an infinite number of points between $a$ and $b$, and this combined with the above suggests $f$ is unbound on $[a,b]$.
To put it more formally, for any $n \in \mathbb{N}$, we can divide the interval $[a,b]$ into a partition of $n$ parts: $P = \{t_0, t_1, t_2,\dots, t_n\}$, with $a = t_0 < t_1 < t_2 < \dots < t_n = b$.
We see then that
$$f(b) \geq f(a) + n\cdot \varepsilon'.$$
Since $\varepsilon' > 0$ and $n$ can be chosen to be arbitrarily large, $f(b)$ must be unbounded, which is absurd.
Our assumption that $f$ is both strictly increasing and discontinuous on all of $[a,b]$ must be wrong.
Therefore, if $f$ is increasing, it must be continuous somewhere in $[a,b]$.
Finally, note that these arguments can be applied to any closed subinterval of $[a,b]$. The set of points at which $f$ is continuous form a "dense set" on $[a,b]$.
$\blacksquare$
Note: The theorem seems to contradict the lemma used to prove it! If we look closely though, it seems ok.

*

*We assumed $f$ was both strictly increasing and completely discontinuous.

*If both of these things are true, there must be a minimum distance between $f$ values.

*If there is a minimum distance between $f$ values, $f$ is not bound on the interval.

*$f$ is bound, therefore, no minimum distance.

*No minimum distance $\implies$ $f$ cannot be both increasing and completely discontinuous.

*If $f$ is definitely increasing then $f$ is continuous somewhere.

 A: The part of the proof you refer to as the "sharkiest" is in fact incorrect. At least, your argument is. This is actually a super key point as you go further into analysis:
The fact that for each pair you can find an $m_i$ that works does NOT imply the existence of an $m'$ that always works. This is faulty logic.
For example, for any $\epsilon>0$ I can always find an integer $n$ so that $1/n<\epsilon$. But there is no $n$ that works for EVERY $\epsilon$. That is, just because there's always a positive integer whose reciprocal is less than any positive real number, that does NOT mean there is an integer whose reciprocal is less than EVERY positive real number. This incorrect logic is the same that you use in your proof.
The end result of your whole proof (as pointed out in the comments) is correct, but your logic, at least there, is wrong.
Edit: how to fix it.
Consider that midpoint $m_1$. Now, either $f(m_1)$ is less than or equal to $k_1=(f(a_1)+f(b_1))/2$ or it's bigger. If it's bigger, consider the interval $I_2=[m_1,b_1]$. Otherwise, consider $I_2=[a_1,m_1]$. Next, set $a_2$ and $b_2$ to be the endpoints of the interval you just considered. Repeat the process inductively. Each time you repeat this process, you cut the maximum possible length of the range of the set $I_n$ in half. Consider $m$ to be a single point in the intersection of all $I_n$. Since they're nested intervals, for any $\epsilon>0$, there's an $I_n$ small enough to contain $m$ and have its range be an interval with range less than $\epsilon$. I've left out some details (such as, what if $m$ is an endpoint of one of the intervals? Could it be, and what do we do if so? How do we even know the intersection is nonempty?). I'll let you fill them in since they can be a good exercise..
A: In your argument, you have a sequence of points $m_i$ and $\epsilon_i \to 0$ so that
$$\tag{1} |x-m_i|<\delta_i \Rightarrow |f(x) - f(m_i)|<\epsilon_i.$$
and from there conclude the existence of $m$ so that
$$|x - m'| < \delta \text{, then } |f(x) - f(m')| < \varepsilon.$$
This is not true. For example, $f: [-1, 1]\to \mathbb R$
$$ f(x)=  \begin{cases} 1& \text{ if } x=0, \\ 0 & \text{ if } x\notin\mathbb Q, \\ x & \text{ otherwise.}\end{cases}.$$
admits such a sequence $\{m_i\}$, (and indeed $m_i \to 0$), but $f$ is nowhere continuous.
Indeed, this is NO function $f: [a,b] \to \mathbb R$ which satisfies
$$ |f(x) -  f(y)| \ge \epsilon'$$
whenever $x, y\in [a, b]$ and $x<y$. To see this: partition $\mathbb R$ into a countable disjoint union
$$ \mathbb R = \bigcup_{q\in \mathbb Z} [q\epsilon', (q+1)\epsilon').$$
then since $[a, b]$ is uncountable, one of $f^{-1}[q\epsilon', (q+1)\epsilon')$ is uncountable and thus has two elements $x<y$. Then $|f(x) - f(y)|< \epsilon'$.
