I am trying to prove that every lindelof regular space is normal, and think I have managed to do so, but my proof seems slightly different to others I have found online. I would appreciate if you could let me know if it works!
Let X a topological space that is lindelof and regular. Let $C,D$ closed subspaces of $X$. By regularity, we can find for each $c \in C$ and $d \in D$, an open $U_c$ and $V_d$ respectively such that
$c \in U_c \subseteq \overline{U_c} \subseteq X \setminus D$,
$d \in V_d \subseteq \overline{V_d} \subseteq X \setminus C$.
Therefore $\{U_c : c \in C\}$ and $\{V_d : d \in D\}$ are open covers of $C$ and $D$ respectively, and so (since $X$ is Lindelof) they have countable subcovers, say $\{U_n : n \in \mathbb{N}\}$ and $\{V_n : n \in \mathbb{N}\}$. (I do realise a tiny bit of work needs to be done to show we can apply $X$ being Lindelof to covers of $C$ and $D$, but I am confident I can do this part)
Now Let $U_n' = U_n \setminus \bigcup_{i=1}^n \overline{V_n}$, and $V_n' = V_n \setminus \bigcup_{i=1}^n \overline{U_n}$.
Then let $U = \bigcup_{i=1}^n U_n'$ and $V = \bigcup_{i=1}^n V_n'$. Then $U$ and $V$ are open, since they are the union of open sets. It is also a simple check (as far as I can tell) that they are disjoint and contain $C$ and $D$ as subsets respectively.
Does my proof work? The most similar proof I have seen constructs instead $U_n' = \bigcup_{i=1}^n U_n \setminus \bigcup_{i=1}^n \overline{V_n}$ and similarly for $V_n'$. Is there any reason for this difference? Does it maybe not matter in this context, but in a different one? Any help is appreciated :)
Your proof works; look at my note here, in particular the first two theorems, so you can see the full details, which are not hard, indeed.