Function everywhere left continuous yet not right continuous everywhere? Is there such a function? I think such a function can only have countable noncontinuous points, maybe I have to use Baire's theorem?
 A: Yes; take a step function and define the value at the left endpoint to be the same as the right endpoint of the previous "step" , i.e., $f(x)=1$ , in $[0,1]; f(x)=2$ on $(1,2]$ , etc. 
A: Everywhere left-continuous function (on $\mathbb{R}$) can have only a countable set of points where it is not right-continuous. See https://mathoverflow.net/questions/148866/is-there-a-function-defined-on-real-numbers-which-is-continuous-from-the-left-b/
A: I am not very satisfied with the answers under this MO post. So let me try to compose a self-contained answer.
So we want to prove that:

Let $ f $ be an arbitrary function on $ (-\infty, +\infty) $ and $ L $ be the set of $ x $ where $ f $ is right continuous but not left continuous. Prove that $ L $ is a countable set.(Exercise 4 on page 7, A Course in Probability Theory by Kai Lai Chung)

Proof:
Consider $ L\cap M_n $, where $ M_n=\{x: O(f;x)>\frac 1n\} $ and $ O(f;x) $ is the oscillation of $ f $ at $ x $. It follows that $ L=\bigcup_{n=1}^\infty (L\cap M_n) $. It is sufficient to prove that $ L\cap M_n $ is countable for each $ n\in\mathbb N $.
Suppose the converse, i.e., $ L\cap M_{n_0} $ is uncountable for some $ n_0\in\mathbb{N} $. We claim that there must exist a point $ P $ in $ L\cap M_{n_0} $ such that $ P $ is a limit point of $ L\cap M_{N_0} $. Otherwise, every point in $ L\cap M_{n_0} $ is sitting in some open interval without points in $ L\cap M_{n_0} $. And we can find a rational number in each open interval. Since rational numbers are countable, it follows that $ L\cap M_{n_0} $ is countable which contradicts our assumption.
Indeed, we can say more. Every uncountable subset $ S $ of $ \mathbb R $ contains all but countably many its limit points. Since every isolated point is contained in a small open interval, which also contains some rationals, such that we can cover the set of all isolated points by a disjoint union of small open intervals. The complement of this set which is the set of all limit points is thus uncountable.
Moreover, we claim that every uncountable subset $ S $ of the real line $ \mathbb{R} $ has at least one point that is two-sided limit point.
We prove by contradiction, suppose $ S $ has no two-sided limit point, then by the previous paragraph we suppose without loss of generality that all the points of $ S $ are left-sided limit points. For every $ p\in S $, we can find a rational $ r_p\in\mathbb Q $ such that $ (p,r_p) $ contains no point in $ S $. Since $ S $ is uncountable, we can find a rational $ r\in\mathbb Q $ such that $T= \{p\in S:r_p=r\} $ is uncountable. This is already a contradiction, since we can pick $ p_1,p_2\in T $(since $ T $ is uncountable) with $ p_1< p_2 $ and there should be no point of $ (p_1,r) $ in $ S $ by construction, but $ p_1<p_2<r $ and $ p_2\in S $.
Now we are ready to conclude that $ L\cap M_{n_0} $ is countable, since by the definition of $ L\cap M_{n_0} $, the oscillation of each point in this set is at least $ 1/n_0 $, while being an uncountable subset of $ \mathbb R $, it contains a two-sided limit point $ P $ by the above argument. This can never happen, since $ f $ is right continuous at $ P $ at the same time. Thus we are done.
Reference:
https://dantopology.wordpress.com/2010/05/29/the-lindelof-property-of-the-real-line/#:~:text=Every%20uncountable%20subset%20has%20a%20limit%20point.&text=Every%20uncountable%20subset%20contains%20one%20of%20its%20limit%20points.&text=hold%20for%20the%20real%20line,line%20has%20the%20Lindelof%20property.
