Could you show me that any function $f:\mathbb{R}\rightarrow \mathbb{R}$ which is right continuous and has left limit is separable? By ‘separable’ I mean that
there exists a countable set $D$ such that for any $t$, there exists $t_n\in D$, $t_n \rightarrow t$ implies $f(t)=\lim_{n \rightarrow \infty}f(t_n)$.
The assertion remains true if “$t_n\rightarrow t$” is replaced by “$t_n \uparrow t$”.
For any real number $t$ there exists a sequence of rationals decreasing to $t$. So take $D$ to be the set of all rationals numbers.
Answer for the modified question: there are at most countable many points $d_1,d_2,...$ at which $f$ is not continuous. Let $D$ be the union of this countable set with the rationals. For $t=d_i$ choose the constant sequence $t_n=t$ for all $n$.