Discontinuities of a monotone function. Let $F : \mathbb{R}\to \mathbb{R}$ be a monotone function then Prove that
 $F$ has countably many discontinuities.
Edit: Monotone function is not only increasing.
My Attempt : 
I know that the set of discontinuities of a monotone function is at most countable.
If I take greatest integer function $[x]$ which is monotone function that has countable set of discontinuities as $\Bbb Z $. But I'm unable to Prove it. Please help me.

Let $D$ denote the set of discontinuity points. For each $x\in D$, the left and right limits differ, and are therefore the endpoints of an interval with non-empty interior. In this manner we obtain a collection of such intervals
$$
\{I_d\colon d\in D\}.
$$
By monotonicity the intervals are disjoint. Choosing one rational number from each interval therefore yields an injection from $D$ into $\mathbb Q$. Hence $D$ is countable.

EDIT: I see there is some confusion on the part of the question asker. The terminology "countable" normally means "has cardinality less than or equal to that of the natural numbers". In particular, that is the only interpretation of that word which makes your question correct. (It is easy to give an example of a monotone function with only a finite number - or even zero - discontinuities.) So if the question asker understands already why there are at most countably many discontinuities, that means there is actually no question left to answer: since "at most countably many" is a synonym for "countably many".
