My motivation for this question is as follows.

Consider the set $D$ of cadlag functions on $(0,1)$ and its subset $D^\uparrow$ of non-decreasing cadlag functions.
Each $f \in D$ has at most countably many points of discontinuity.
One can equip $D$ with the Skorokhod $M_1$ topology which turns both $D$ and its subspace $D^\uparrow$ into Polish spaces.
Convergence $f_n \to f$ in $D^\uparrow$ under the $M_1$-topology is equivalent to $f_n(x) \to f(x)$ for each continuity point $x$ of $f$.

On $D$ such a pointwise characterization of the $M_1$-convergence does not hold.
I was asking myself, if there is some relation between such a pointwise convergence at points of continuity to the $M_1$-topology.

Let us start more abstractly and consider a topological space $X$ ($X = \mathbb{R}$ is enough for our considerations).
Let $F \subseteq X^{[0,1]}$ be some function space and define convergence as follows:
$f_n \to' f$ if and only if there exists a dense subset $D \subseteq X$ such that $f_n(x) \to f(x)$ for all $x \in D$.
Before speaking about the induced (sequential) topology, it is necessary to check whether this convergence defines a limit space (synonymously a subsequential space or an $L$-space),
i.e. one has to check the Kuratowski axioms (without uniqueness of limit) which are:

(i) for constant sequence $f_n = f$ we have $f_n \to' f$

(ii) if $f_n \to' f$ then $f_{n_k} \to' f$ for every subsequence $f_{n_k}$

(iii) (subsequence principle) if every subsequence $f_{n_k}$ has a subsubsequence $f_{n_{k_l}}$ with $f_{n_{k_l}} \to' f$ then $f_n \to' f$.

Properties (i) and (ii) do trivially hold but  I can't show (iii) in such a generality and I think that it does not hold for $\to'$.

If it is wrong in general, what about the special case of such a definition of convergence (or even hopefully of a (sequential) topology)
for the space $F = D$ of cadlag functions as above?

What happens if we restrict to define $\to'$ to be pointwise convergence outside a countable set (for $X = \mathbb{R}$ the complement of a countable set is dense).

(Maybe there is some similarity to the fact that on measure spaces convergence a.e. does not define a topological space but it does at least define a limit space.)