Let $(E,d)$ be a metric space and $x\colon[0,\infty)\to E$ such that $x(0)=0$ and $$x(t\pm):=\lim_{s\to t\pm}x(s)$$ exists for all $t\ge0$ ($x(0-):=x(0)$).
Let $t\ge0$ and $r>0$ with $d(x(t+),x(t-)\ge r$.
How can we show that there is a $\delta>0$ with $$\forall t-\delta<s\le t\le u<t+\delta:d(x(s),x(u))\geq r?\tag1$$
This conclusion seems to be claimed in the second paragraph of this answer (but please feel free to fix the claim, if necessary).
Given $\varepsilon>0$, we can clearly find a $\delta>0$ such that $$d(x(s),x(t-))<\frac\varepsilon2\;\;\;\text{for all }s\in(t-\delta,t)\tag2$$ and $$d(x(t+),x(u))<\frac\varepsilon2\;\;\;\text{for all }u\in(t,t+\delta)\tag3,$$ but this only yields $$r\le d(x(t+),x(u))+d(x(u),x(s))+d(x(s),x(t-))\le\varepsilon+d(x(u),x(s))\tag4$$ for all $t-\delta<s<t<u<t+\delta$ ...
The conclusion does not hold. As a counterexample you can take the process $$ x(t)= \begin{cases} 1-t & \text{if }t\in[0,1), \\ 2-t & \text{if }t\in[1,\infty). \end{cases} $$ This process has a jump at time $1$ with $\Delta x(1)=1$. Note however, that for any $s\in(0,1)$ and $u\in(1,2)$ we have that $x(s),x(u)\in(0,1)$. Hence $\lvert x(s)-x(u) \rvert<1$.
To "fix" this, you should replace "$\geq r$" with "$>r$". Intuitively you can think of the following. If $(a_n)$ is a sequence in $\mathbb{R}$ which converges to some limit $a$, then the conclusion $$ a\geq r\Rightarrow a_n\geq r\text{ for all } n \text{ large enough} $$ is generally false (take $a_n=r-1/n$). Instead the we have that $$ a> r\Rightarrow a_n> r\text{ for all } n \text{ large enough}. $$