How to prove that every Paracompact space with the Suslin property is Lindelof This question was asked a few years ago and a proof was given here https://math.stackexchange.com/a/190147/235467. 
However, in this proof it states that paracompactness implies the existence of a $\sigma$-discrete open refinement for any open cover. However, paracompactness only implies a locally finite open refinement. These are not the same, in general. I think I stumbled across something that stated that adding regular to the assumptions gives you the $\sigma$-discrete refinement but I can't seem to find that particular resource again. If that is the case, then my concerns are resolved as Paracompact Hausdorf implies regularity and normality. Any help would be greatly appreciated! 
If regularity fixes my concern, could you please leave a hint as to how to prove it? 
 A: Yes, a $T_3$-space is paracompact iff every open cover of it has a $\sigma$-discrete open refinement. In the course of this answer I proved that if $X$ is a paracompact Hausdorff space, then every open cover of $X$ has an open barycentric refinement, and in the course of this answer I proved that a barycentric open refinement of a barycentric open refinement of an open cover is an open star refinement of the original open cover. Thus, it suffices to show that if every open cover of $X$ has an open star refinement, then every open cover of $X$ has a $\sigma$-discrete open refinement. This is Lemma $\mathbf{5.1.16}$ in R. Engelking, General Topology.
Assume that every open cover of $X$ has an open star refinement, and let $\mathscr{U}=\{U_\xi:\xi<\kappa\}$ be an open cover of $X$. Let $\mathscr{U}_0=\mathscr{U}$, and for $n\in\Bbb N$ let $\mathscr{U}_{n+1}$ be an open star refinement of $\mathscr{U}_n$. For $\xi<\kappa$ and $n\in\Bbb Z^+$ let
$$U_\xi(n)=\{x\in X:\operatorname{st}(V,\mathscr{U}_n)\subseteq U_\xi\text{ for some open nbhd }V\text{ of }x\}\;.$$
Then for each $n\in\Bbb Z^+$, $\{U_\xi(n):\xi<\kappa\}$ is an open refinement of $\mathscr{U}$ covering $X$.
Suppose that $n\in\Bbb Z^+$, $\xi<\kappa$, and $x\in U_\xi(n)$. Suppose further that $x\in U\in\mathscr{U}_{n+1}$. $\mathscr{U}_{n+1}$ star refines $\mathscr{U}_n$, so there is a $W\in\mathscr{U}_n$ such that $\operatorname{st}(U,\mathscr{U}_{n+1})\subseteq W\subseteq\operatorname{st}(x,\mathscr{U}_n)\subseteq U_\xi$, and by definition $U\subseteq U_\xi(n+1)$. Thus, $\operatorname{st}(x,\mathscr{U}_{n+1})\subseteq U_\xi(n+1)$.
Now for each $\eta<\kappa$ and $n\in\Bbb Z^+$ let
$$V_\eta(n)=U_\eta(n)\setminus\operatorname{cl}\bigcup_{\xi<\eta}U_\xi(n+1)\;,$$
and let $\mathscr{V}_n=\{V_\xi(n):\xi<\kappa\}$; clearly each $\mathscr{V}_n$ is an open refinement of $\mathscr{U}$. I claim that each $\mathscr{V}_n$ is a discrete collection, and that $\mathscr{V}=\bigcup_{n\in\Bbb Z^+}\mathscr{V}_n$ covers $X$.


*

*Fix $n\in\Bbb Z^+$, and suppose that $\xi<\eta<\kappa$; then $V_\eta(n)\subseteq X\setminus U_\xi(n+1)$, and $V_\xi(n)\subseteq U_\xi(n)$. Suppose that $x\in V_\xi(n)$ and $y\in V_\eta(n)$. Then $\operatorname{st}(x,\mathscr{U}_{n+1})\subseteq U_\xi(n+1)$, while $y\notin U_\xi(n+1)$, so no member of $\mathscr{U}_{n+1}$ contains both $x$ and $y$, and it follows that $\mathscr{V}_n$ is discrete.

*Let $x\in X$. Let $\eta=\min\{\xi<\kappa:x\in U_\xi(n)\text{ for some }n\in\Bbb Z^+\}$; this exists because each $\{U_\xi(n):\xi<\kappa\}$ for $n\in\Bbb Z^+$ is an open cover of $X$. Fix $n\in\Bbb Z^+$ such that $x\in U_\eta(n)$. If $\xi<\eta$, then $x\notin U_\xi(n+2)$. Suppose that $y\in\bigcup_{\xi<\eta}U_\xi(n+1)$; then $\operatorname{st}(y,\mathscr{U}_{n+2})\subseteq U_\xi(n+2)$, so $x\notin\operatorname{st}(y,\mathscr{U}_{n+2})$, and hence $y\notin\operatorname{st}(x,\mathscr{U}_{n+2})$. It follows that $\operatorname{st}(x,\mathscr{U}_{n+2})$ is an open nbhd of $x$ disjoint from $\bigcup_{\xi<\eta}U_\xi(n+1)$ and hence that $$x\in U_\eta(n)\setminus\operatorname{cl}\bigcup_{\xi<\eta}U_\xi(n+1)=V_\eta(n)\;.$$ Thus, $\mathscr{V}$ covers $X$.
