When is: 'X paracompact iff X has $\sigma$-locally finite refinement' true? I am reading my course notes about paracompact spaces and there is the following result:

$X$ is paracompact iff every open cover of $X$ has a $\sigma$-locally
  finite open refinement.

However the proof uses the following result from a problem sheet we did:

For a regular space, the following are equivalent,
  
  
*
  
*X is paracompact
  
*Every open cover of $X$ has a locally finite refinement
  
*Every open cover of $X$ has a locally finite closed refinement
  

However the original theorem in the notes does not include any seperation assumption. So is the first result only valid for regular spaces?
EDIT
Looking at Willards 'General Topology', theorem 20.12 says 

Every $F_\sigma$ subset of a paracompact space is paracompact. 

The proof of this result shows that every open cover of a $F_\sigma$ subset has a $\sigma$-locally finite refinement. 
Hence is the implication 

$X$  has a $\sigma$-locally finite open refinement $\implies X$ is
  paracompact

true without the regularity property, but the converse needs the regularity? 
 A: I use regularity in the proof of $2 \implies 3$ of from your equivalence (from the problem sheet), but not in the other steps. See my note on this here, if you want to check my proof. It's 3 to 4 there as I add another equivalence with "every open cover has a $\sigma$-locally finite open refinement" in that lemma; so we have it all in one go, so the formulation I chose was the equivalence (for regular spaces) of 


*

*$X$ paracompact (every open cover has a locally finite open refinement).

*Every open cover has a $\sigma$-locally finite open refinement.

*Every open cover has a locally finite refinement (i.e. a cover but not necessarily open).

*Every open cover has a locally finite closed refinement.


*

*$1 \implies 2$ is trivial and needs nothing, just like $1 \implies 3$.

*$2 \implies 3$ needs no regularity in the proof I gave.

*For $3 \implies 4$, I do use the regularity.

*$4 \implies 1$ also doesn't use it. 



Your first result you state is the equivalence of 1 and 2 and I see no path here from 2 to 1 that avoids 3 to 4 somehow... So regularity seems needed.
An example: if $X$ is a Lindelöf Hausdorff, non-regular space, then $X$ obeys the property that every open cover has a $\sigma$-locally finite open refinement (a countable subcover will do that easily) but such an $X$ is not paracompact as a paracompact Hausdorff space is both regular and normal, as is well-known.
Such spaces have been found in abundance (see $\pi$-base e.g., the Arens square and the irational slope topology are the most "famous")
The $F_\sigma$ subset result will hold because Willard (as Engelking) defines a paracompact space to be Hausdorff, and such a space is thus normal and regular, and so the $F_\sigma$ subspace is regular too and the lemma applies to it.
