Show that if $x$ is càdlàg, $\left\{t\in[a,b]:d\left(x(t),x(t-)\right)>c\right\}$ is finite Let $(E,d)$ be a metric space and $x:[0,\infty)\to E$ be càdlàg. Moreover, let $x(0-):=x(0)$ and $$x(t-):=\lim_{s\to t-}x(s)\;\;\;\text{for }t>0.$$

Let $b>a\ge0$, $c>0$ and $$I:=\left\{t\in[a,b]:d\left(x(t),x(t-)\right)>c\right\}.$$ I want to show that $|I|\in\mathbb N_0$.

I've got a concrete idea, but I'm struggling to finish my argumentation: Assume $|I|\not\in\mathbb N_0$. Then there is a $(t_n)_{n\in\mathbb N}\subseteq I$ with $$t_n<t_{n+1}\;\;\;\text{for all }n\in\mathbb N\tag1.$$ Since $[a,b]$ is sequentially compact, $\left(t_{n_k}\right)_{k\in\mathbb N}$ is convergent for some increasing $(n_k)_{k\in\mathbb N}\subseteq\mathbb N$.

My idea is now to obtain a contradiction by showing that $\left(t_{n_k}\right)_{k\in\mathbb N}$ cannot be Cauchy. In order to do this, we clearly want to use that $\left(t_{n_k}\right)_{k\in\mathbb N}\subseteq I$ implies $$c<d\left(x\left({t_{n_k}}\right),x\left({t_{n_k}}-\right)\right)\le d\left(x\left({t_{n_k}}\right),x\left({t_{n_l}}\right)\right)+d\left(x\left({t_{n_l}}\right),x\left({t_{n_k}}-\right)\right)\tag2$$ for all $k,l\in\mathbb N$. If we now could show that there is a $k_0\in\mathbb N$ with $$d\left(x\left({t_{n_l}}\right),x\left({t_{n_{k_0}}}-\right)\right)<\frac c2\tag3\;\;\;\text{for all }l\ge k,$$ we would be done. But how can we show this?

I've tried several things, but they don't seem to be useful. For example, if $t\in[a,b]$ is the limit of $\left(t_{n_k}\right)_{k\in\mathbb N}$, we may use that $x$ has a left-limit at $t$ to conclude that there is a $\delta>0$ with $$d\left(x(s),x(t-)\right)<\frac c2\;\;\;\text{for all }s\in(t-\delta,t)\tag4.$$ Now, there is a $k_0\in\mathbb N$ with $$t-\delta<t_{n_k}<t\tag5\;\;\;\text{for all }k\ge k_0.$$ However, this is not enough to conclude ...
 A: Let us fix an interval $[a, b]$ and a number $c > 0$. We want to show that $d(x(u-), x(u)) > c$ is possible for at most finitely many $u \in [a, b]$.
It is possible to prove this by assuming that there is a sequence of pairwise distinct $(u_n)$ in $[a, b]$ with $d(x(u_n-), x(u_n)) > c$ for all $n$, which must have an accumulation point, and then derive a contradiction. But I prefer to use a compactness argument:
For $s \in [a, b]$ there exist a $\delta_s > 0$ such that
$$
  d(x(t), x(s)) < \frac c 2 \text { for } s < t < s + \delta_s
$$
and
$$
  d(x(t), x(s-)) < \frac c 2 \text { for } s-\delta_x < t < s \, .
$$
The first estimate follows from the right-continuity at $s$, and the second from the existence of a left-limit at $s$. Then
$$
 d(x(t), x(u)) \le d(x(t), x(s))+d(x(s), x(u)) < c
$$
for $s < t < u < s+\delta_s$, and the same inequality holds for $s -\delta_s< t < u < s$. Taking the limit $t \to u$ we conclude that
$$
 d(x(u-), x(u)) \le c \text{ for } u \in I_s \setminus \{ s \} 
$$
with $I_s = (s-\delta_s, s+\delta_s)$.
Now $\{ I_s \mid s \in [a, b] \}$ is an open covering of the compact set $[a, b]$. It follows that there are finitely many $s_1, \ldots, s_n$ such that
$$
 [a, b] \subset \bigcup_{n=1}^n I_{s_n} 
$$
and therefore $d(x(u-), x(u)) \le c$ for all $x \in [a, b]$ except possibly at the points $s_1, \ldots, s_n$. This concludes the proof.
