TITLE: How to prove that càdlàg (RCLL) functions on $[0,1]$ are bounded?
QUESTION [9 upvotes]: While studying the space $\mathbb{D}[0,1]$ of right continuous functions with left hand limits (i.e. càdlàg functions) on $[0,1]$, I came across the following theorem:

Theorem. If $f$ is càdlàg on $[0,1]$, it is bounded.

My proof attempt:
I am aware that if a function has both left and right hand limits on $[0,1]$, then the set of discontinuities is at most countable. Hence I tackled this in two parts, one where the discontinuities are finite, the other infinite.
I got the finite discontinuity one. But I am stuck at the infinite discontinuity part. My guess is that there is something special about the countable discontinuities (e.g. they cannot be dense in $[0,1]$) and somewhere, I'll have to use the sequential compactness property to get a contradiction, but I am unable to collect my ideas. 
I request any starting hints on this. A sketch of the proof would also be appreciated.

REPLY [7 votes]: Billigsley gives an excellent one-line proof. I am repeating it here. First note that this Lemma is true :
For every $\epsilon >0$,   $ \exists$ partition $ 0=t_0<t_1<\ldots,t_k=1$ such that for the set $S_i = [t_i,t_{i+1})$, we have   $\sup_{s,t \in S_i} |f(s)-f(t)| <\epsilon$ for all $i$.
This Lemma is easily proved. Once Lemma is seen to be true, choose an $\epsilon >0$. Then the bound on $f$ is simply $(\sum_{l=1}^{k} J_{l}  ) + \epsilon \times k$ (first term is the amount of jumps which occur at the $k$ points which were obtained using the Lemma, while the second term is the increment of $f$ within these intervals.