The answer by @KaviRamaMurthy is elegant and ingenious since his $\ F\ $ is monotone (increasing) -- thus, while the construction of $\ F\ $ is a bit involved, the proof is instant.
The following is perhaps more straightforward(?); I'll make it general so that it's even easier to see what's going on (altogether not too much):
Let $\ A\subseteq X\subseteq\mathbb R,\ $ where $\ X\setminus A\ $ is dense in $\ X,\ $ and let
$\ g:A\rightarrow\mathbb R\setminus\{0\}\ $ be such that
$$ s\ :=\ \sum_{a\in A}\,|g(a)|\ <\ \infty $$
(it follows that $\ A\ $ is countable). Define
$\ G:X\rightarrow\mathbb R\ $
as the extension of $\ g\ $
by $\ 0,\ $ i.e. $\ \forall_{x\in X\setminus A}\, G(x)=0.\ $
It follows that $\ G\ $ is continuous at every $\ x\in X\setminus A,\ $ and that the total variation is
$\ \le\ 2\cdot s$.
Since $\ X\setminus A\ $ is dense in $\ X,\ $ function $\ G\ $
is discontinuous at every $\ a\in A.$