Set of discontinuity of monotone function is countable

The usual definition of a series of nonnegative terms is as the supremum of the sums over finite subsets of the index set, $$\sum_{i\in I} x_i=\sup\biggl\{\sum_{j\in J}x_j:J\subseteq I\mbox{ is finite}\biggr\}.$$ (Note this definition does not quite work in general for series of positive and negative terms.)

The point then is that is $a< x<y< b$ and $F\!:[a,b]\to\mathbb R$ is increasing, then $$ F(a)\le F(x^-)\le F(x^+)\le F(y^-)\le F(b). $$ It follows from this that if $J\subset (a,b)$ is a finite set of discontinuities of $F$, then $$ \sum_{z\in J}F(z^+)-F(z^-)\le F(b)-F(a) $$ and therefore the same holds for the sum $S $ over all discontinuity points of $F$ in $(a,b)$. One way to think about this is to note that the (finitely many) intervals $(F (z^-),F (z^+)) $, $z\in J $, are pairwise disjoint and are all contained in $(F (a), F (b))$, so the sum of their lengths is at most the length of the whole interval.

Now, if $x$ as above is actually a discontinuity point of $F$, then the inequality $F(x^-)<F(x^+)$ is strict. It follows from this that the sum I called $S $ above is over a countable index set, since any series of uncountably many positive terms diverges.

Note that I did not include $a$ or $b$ in the above, but of course this does not change the conclusion that the set of discontinuities of $F$ in $[a,b]$ is countable. If we include them, then we need to use the convention that $F(a^-):=F(a)$ and $F(b^+):=F(b)$. With this convention the same conclusion (with the same bound) holds: If $D$ is the set of discontinuity points of $F$ in $[a,b]$, then $$ S=\sum_{x\in D}F(x^+)-F(x^-)\le F(b)-F(a). $$

Naturally, the same conclusion (that $D$ is actually countable) follows if $F$ is defined not just in a finite interval but on an unbounded one (even on $\mathbb R$), by noting that $\mathbb R=\bigcup_{N\in\mathbb N}[-N,N]$.