I have spent a week on this problem without making any progress. I am not sure what I am missing. The problem is as follows
For any arbitrary function $f:\mathbb{R}\rightarrow \mathbb{R}$ the set of points $X = \{x_o : \limsup_{x \rightarrow x_o^-} f(x) > \limsup_{x \rightarrow x_o^+ } f(x) \}$ is countable.
I get the fact that this set if uncountable has to have a limit point. Unlike showing removable discontinuities have to be countable, I cannot seem to find what the contradiction is. I get that the $\limsup$ is a limit point of the set of local supremums of the function. However, I cannot seem to make the case for a contradiction. I would also appreciate a good example with a countable set where the above condition is true. There is a gap in my understanding that is preventing me from seeing it.
For each point $x_0\in X$, we find a rational number $q_1(x_0)$ such that $$\tag1 \limsup_{x\to x_0^+}f(x)<q_1(x_0)<\limsup_{x\to x_0^-}f(x).$$ Also, this means that $\sup_{x\in(x_0,x_0+\epsilon)}f(x)<q_1(x_0)$ (and hence also $f(x)<q_1(x_0)$ for $x_0< x<x_0+\epsilon$) if $\epsilon>0$ is small enough. This lets us pick a second rational number $q_2(x_0)>0$ with $f(x)<q_1(x_0)$ for all $x$ with $0<x-x_0<q_2(x_0)$.
Assume $X$ is uncountable. As there are only countably many pairs of rationals, some set of the form $$\tag2 X_{a_1,a_2}=\{\,x\in X\mid q_1(x)=a_1, q_2(x)=a_2\,\}$$ with $a_1,a_2\in\Bbb Q$ is uncountable and has a limit point $b$. Let $\{x_n\}_n$ be non-constant sequence of points in $X_{a_1,a_2}$ with $x_n\to b$. As this is a Cauchy sequence, we have $|x_n-x_m|<a_2$ for all large $n,m$. Then if wlog. $x_n<x_m$, we conclude that $f(x)<a_1$ for all $x\in(x_n,x_m)$. But this implies $$a_1=q_1(x_m)<\limsup_{x\to x_m^-}f(x)\le a_1,$$ contradiction.