I found these terms undefined, and I've looked around and can't find the definition. It does not appear in the handbook of set-theoretic topology or the handbook of set-theory, and it does appear on the handbook of the history of topology, but undefined.
A screenshot for context may be of use.
Thanks for any help.

An "éspace (L)" (L for limit, I think), as Fréchet names it, is a space $X$ with a notion of convergence $C$ of sequences, so we can say that a sequence $x_n \to x$ under $C$, and this notion tells us which sequences do, you could in a modern fashion say that it is a function $C: X^\omega \times X \to \{\mathbf{true}, \mathbf{false}\}$, where $C((x_n), x) = \mathbf{true}$ exactly when $x_n \to_C x$, and $x_n \not\to_C x$ otherwise, that satisfies two basic axioms:
In such a space-$L$, we can define a set to be open by
One can check that this defines a topology in the usual sense.
This notion has been studied quite widely in the old days of topology, and even in general topology today, Fréchet(-Urysohn) spaces are topological spaces that are rather nicely behaved w.r.t. sequence convergence. Engelking (General Topology) has an exercise on them at the end of chapter 1.
The space-$H$ is more obscure. I found this in the survey paper by Angus E. Taylor, "A Study of Maurice Fréchet: II. Mainly about his Work on General Topology, 1909-1928", which I found here and which looks like it comes from some journal.
First there is a notion of "éspaces (V)" (V for voisinage, neighbourhood in French): a set $X$ where every point has a set of neighbourhoods of that point.
The base axioms for these "Fréchet-$V$ spaces" are
Such a space is then called a classe (H) (in honour of A.R Hedrick, an American topologist), if it also obeys:
These $V$ spaces and $H$ spaces also have a notion of topology associated with them.
Note that your paper by Jones was written in a time that different competing axiomisations of topology were around and in use. Defining everything in terms of open sets was already one of them, but not the dominant one yet. Even the set theory notations are still old-fashioned in modern eyes ($+$ for union etc.).
You could mostly replace these spaces-$L$ and spaces-$H$ by topological space here, I think, except where there is reasoning with sequences, e.g.
Added after comments
To illucidate the argument given by Jones (so we don't use extended characteristic functions) : For each proper non-empty subset $J$ of $M$ (which is closed in $S$, as well as $M\setminus J$) we can by normality of $S$ find open sets $D_J$ and $E_J$ of $S$ such that $D_J \cap E_J = \emptyset$ and $J \subseteq D_J$ and $M\setminus J \subseteq E_J$. Then if $J \neq K$, we can WLOG (as everything is symmetric) assume that $J \setminus K \neq \emptyset$. Let's denote by $Z_J$ the intersection $Z \cap D_J$
But then, as $J \setminus K \subseteq D_J \cap E_K$ so the latter set is non-empty open in $S$, so its intersection with the dense set $Z$ is non-empty.
We then see that $\emptyset \neq Z \cap D_J \cap E_K \subseteq Z_J \setminus D_K \subseteq Z_J \setminus Z_K$, so that $Z_J \neq Z_K$. So indeed distinct $J$ give rise to distinct $Z \cap D_J$, as claimed. No special things needed, just normality.
Alternatively, we can note that $D_J \cap M = J = \overline{D_J} \cap M$, and because $Z$ is dense and $D_J$ is open, we have that $\overline{D_J} =\overline{Z \cap D_J}$, which also implies that distinct $J$ and $K$ have distinct $Z_J$ and $Z_K$. This is probably what the author had in mind.
Theorem 2 is an almost immediate corollary of Theorem 1, if you use that $X$ is completely normal iff every (open) subspace of $X$ is normal:
If $M$ were discrete in $S$, $M$ would be closed and discrete in $S \setminus M'$, and $S$ separable implies $S \setminus M'$ separable (separability is open hereditary), and $S$ completely normal implies $S \setminus M'$ is normal. This then contradicts Theorem 1. So $M$ is not discrete (i.e. $M$ contains a limit point of itself), so it's not (for me) an adaptation of the proof, but just a straightforward consequence. 3 and 4 are just reformulations: the proof of 1 there really shows that $2^{|M|} \le 2^{\aleph_0}$, because $J \to Z_J$ is 1-1, and if $M$ is uncountable, $|M| \ge \aleph_1$ and so $2^{\aleph_1} \le 2^{|M|} \le 2^{\aleph_0}$. So if we assume $2^{\aleph_0} < 2^{\aleph_1}$, $M$ cannot be uncountable. etc.