It would be appreciated if you could review my proof for correctness. Thanks!
Problem:
Show that every regular Lindelöf space is normal
Proof:
Let X be a regular Lindelöf space.
Let $A$, $B$ be closed sets in $X$ that are disjoint.
Note that since $X$ is regular that implies that one point sets are closed.
I assert the following lemma:
If $A$ is a closed set in a Lindelöf space, then $A$ is Lindelöf.
Proof of the above:
Let $A$ be a closed set in $X$, where $X$ is Lindelöf.
Then let $\mathcal{C}$ be some open covering of $A$ by open sets in $X$. Then $\mathcal{C} \cup \{X – A\}$ is an open covering of $X$ since $X – A$ is open by that fact that $A$ is closed.
Since $X$ is Lindelöf there is a countable subcover which may or may not include $(X – A)$. If it does, simply remove the set $X – A$ and we are left with a countable open cover of $A$ as a countable subset of our original set $\mathcal{C}$. Hence $A$ is Lindelöf.
Now to the proof for the main statement:
Since $X$ is regular and $B$ is closed and $A$ is disjoint from $B$, there is then for each $x \in A$ an open set $U_x$ containing $x$ such that $U$ is disjoint from $B$. By regularity we can find an open set $V_x$ such that $\overline{V_x} \subseteq U_x$. We can do this same operation for any point $x \in A$. Do the similar thing for $B$ w.r.t. $A$.
After doing so, we arrive at two open coverings for $A$ and $B$. Since $A$ and $B$ are closed sets in a Lindelöf space we know that $A$ and $B$ are themselves Lindelöf. Then there are open subcovering of those covers of $A$ and $B$, call these open covers $\mathcal{W}=\{W_n: n \in \Bbb N\}$ and $\mathcal{Z}=\{Z_n: n \in \Bbb N\}$ for $A$ and $B$ respectively. We can now define the open sets $W_n$ and $Z_n$, $n \in \Bbb N$ as:
$$W'_n = W_n – \bigcup_{i=1}^n \overline{Z}_i$$
and
$$Z'_n = Z_n – \bigcup_{i=1}^n \overline{W}_i$$
Then consider the open sets $$W' = \bigcup_n W'_n,\, Z' = \bigcup_n Z'_n$$ Then $W'$ and $Z'$ are disjoint since suppose some $y\in W'$ and $y\in Z'$. Then we have for $i \ge j$ that $y \in W_i$ but $y \notin Z_j$ for any $j \le i$. But by the definition of $Z'_i$ we have also that $y \in Z'_j$ for some $j \le i$, hence we have a contradiction. The case where $i \le j$ works similarly.
Hence we have obtained two disjoint open sets $W'$ and $Z'$ that contain the closed sets $A$ and $B$. Hence $X$ is normal.
Best Answer
The proof is in essence correct, but the disjointness argument could be slightly more to the point: suppose we had $x \in W' \cap Z'$, so $x \in W'_i$ for some $i$ and $x \in Z'_j$ for some $j$. Fix these $i$ and $j$, then we have two cases:
If $i \ge j$ then we know $x \in W'_i$, so $x \in W_i$ and $x \notin \overline{Z_j}$, so $x \notin Z_j$, so $x \notin Z'_j$ contradiction.
If $j \ge i$ we know that $x \in Z'_j$ so $x \notin \overline{W_i}$, so $x \notin W_i$, so $x \notin W'_i$, contradiction.
So in either case we have a contradiction and so $W'$ and $Z'$ must be disjoint.
You also do not give an argument why $A \subseteq \bigcup_n W'_n$ and likewise $B \subseteq \bigcup_n Z'_n$, but this is rather easy: $A$ is covered by the $W_n$ to begin with and all $Z_n$ were chosen so that $A \cap \overline{Z_n} = \emptyset$ which implies $A \subseteq X - \bigcup_n \overline{Z_n}$ (but the latter set is not necessarily closed so we cannot use it and we have to resort to finite unions to subtract, which we can, as we saw). The sets $W'_n$ and $Z'_n$ are also open as the difference between an open and a closed set, so indeed $W'$ and $Z'$ are open.