If $F,\varphi :[a,b] \rightarrow \mathbb{R}$ with F continuous, $\varphi$ strictly increasing and $(F+\varphi)(a)>(F+\varphi)(b)$ is $F+\varphi$ decreasing on some interval in $(a,b)$?
This is clearly true if $\varphi$ has finitely many points of discontinuity in the interval but I am unsure if this statement is true if there are infinitely many points of discontinuity. I have a feeling its not true but can't show it.
No, not necessarily. The function $\varphi$ may have a jump discontinuity at each rational number in $(a,b)$, and since $F$ is continuous, $F+\varphi$ will have exactly the same jump discontinuities, jumping up at each rational. With some details, given any interval $(p,q)$ take a rational $r\in(p,q)$, then $\displaystyle\lim\limits_{x\to r^-}(F+\varphi)(x) - \lim\limits_{x\to r^+}(F+\varphi)(x)=\displaystyle\lim\limits_{x\to r^-}\varphi(x) - \lim\limits_{x\to r^+}\varphi(x)<0$ so there are $s,t$ near $r$ with $s<r<t$ and $(F+\varphi)(s)-(F+\varphi)(t)<0$, that is $(F+\varphi)(s)<(F+\varphi)(t)$.
For the construction of such a function $\varphi$ that $\varphi$ could have a jump (up) discontinuity at each rational, or more generally at each element of a given countable set, see Construct a monotone function which has countably many discontinuities
