An example of topological space which $T_5$ but not $T_6$ Definition: A space $X$ is said to be completely normal ($T_5$- space) if every subspace of $X$ is normal. 
Definition: A space $X$ is said to be perfectly normal ($T_6$ - space) if $X$ is normal and if every closed set in $X$ is a $G_{\delta}$ set in $X$.
I was able to show that $T_6$-space is $T_5$-space.
However I am trying to consider an example of a space which $T_5$ but not $T_6$. I considered the space $\overline{S}_{\Omega}={S}_{\Omega}\cup \{\Omega\}$.
I was able to show that this space is not $T_6$ because $\{\Omega\}$ is closed set but is not $G_{\delta}$ set but $\overline{S}_{\Omega}$ is normal space by Theorem 32.4 from Munkres book.

Theorem 32.4. Every well-ordered set $X$ is normal in the order
  topology.

Question:  However I have difficulties showing that $\overline{S}_{\Omega}$ is $T_5$-space. Let's take any subspace $X$ of $\overline{S}_{\Omega}$. We consider $X$  with the subspace topology inherited from $\overline{S}_{\Omega}$. However subspace topology on $X$ may not be the same as the order topology on $X$ (if $X$ was convex then they should be equal). So in this case I cannot apply theorem 32.4 in order to conclude that $X$ is normal subspace.
I would be very thankful if anyone can explain it to me in details, please!
 A: You could read about so-called linearly ordered topological spaces, abbreviated LOTS, also known as orderable spaces, and, on the other hand, generalized ordered spaces, abbreviated GO-spaces, and also known as sub-orderable spaces. 
A LOTS is a linear order with the open interval topology (also called the order topology). 
A GO-space is a linear order with a topology (at least as strong as the order topology, and hence T$_2$) having a basis consisting of order-convex sets. A standard example is the Sorgenfrey line, which as a set is the real line, with a basis consisting of all sets of the form $[a,b)$ with $a<b$. 
It turns out that a space is a GO-space (explaining the terminology sub-orderable) if and only is it is a (topological) subspace of an ordered space. 
For example, let $H=\Bbb R\times\{0,1\}$ with the lexicographic-order (i.e. dictionary order). It has a topology with basis all open intervals, so it is orderable, or LOTS. 
Let $S=\Bbb R\times\{1\}$, then as a topological subspace it is homeomorphic to the Sorgenfrey line, since every interval $[\langle x,1\rangle,\langle y,1\rangle)$ (with $x<y$) is an open set of $S$, because it is the intersection with $S$ of the interval 
$(\langle x,0\rangle,\langle y,1\rangle)$, and the latter is open in $H$. 
Alternatively, given any GO-space, one may add more points and make the given space a subspace of a LOTS. See Theorem 2.9 in the following paper available online. 
A very good introduction to these topics is :
On generalized ordered spaces
D. J. Lutzer
https://eudml.org/doc/268514
http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.desklight-14cae174-9945-42a3-92c2-9d7ae5db4d00/c/rm89_01.pdf
In particular, your question is answered by Proposition 4.1 in the above paper (but refers to other papers), which states that every GO-space is collectionwise normal 
(which is a stronger form of normality).  
Perhaps look at the following two papers for a complete proof. 
A direct proof that a linearly ordered space is hereditarily collectionwise normal,
Lynn A. Steen
Proc. Amer. Math. Soc. 24 (1970), 727-728.
https://www.ams.org/journals/proc/1970-024-04/S0002-9939-1970-0257985-7/
https://www.ams.org/journals/proc/1970-024-04/S0002-9939-1970-0257985-7/S0002-9939-1970-0257985-7.pdf 
and
A SIMPLE PROOF THAT A LINEARLY ORDERED SPACE IS HEREDITARILY 
AND COMPLETELY COLLECTIONWISE NORMAL,
F.S. CATER,
ROCKY MOUNTAIN JOURNAL OF MATHEMATICS, Volume 36, Number 4, 2006
https://projecteuclid.org/download/pdf_1/euclid.rmjm/1181069408
Alternatively, the weaker result that every sub-orderable space is normal is enough to answer your question, and has an easier proof. Theorem 2.9 (a)$\iff$(b) in Lutzer's paper states that a space is a GO-space if and only if it is a closed (topological) subspace of some (suitable) orderable space. Every orderable space is normal (even if Munkres only does it for well-ordered spaces only), and normality is hereditary with respect to closed subspaces. 
I remember reading the theorem that every GO-space is collectionwise normal from a book by Nagata, and the proof there was well-written. (Perhaps you could try to prove this result, using the existence of a basis of order-convex sets.) I don't have the book in front of me, but I could google a reference. 
Modern general topology,
J.-I. Nagata
Elsevier, 1985, 521 pages.
https://books.google.com/books/about/Modern_General_Topology.html?id=ecvd8dCAQp0C 
(The result is also given in Engelking's General topology book, but perhaps only in the exercises.) 
I just looked at Steen's paper again, it seems short and clear, and has a complete proof that every LOTS is hereditarily collectionwise normal, which is (more than) what you need. Hereditarily collectionwise normal means every (topological) subspace is collectiowise normal (which in turn means that not just any two disjoint closed sets, but every discrete collection of closed sets can be separated by disjoint family of open sets). Steen also comments on the difference between the inherited topology and the topology generated by the inherited linear order, and does all that in just two pages. 
