A non-compact $ T_{2} $ space which is not a k - space A topological space is called k - space  provided it has the property that any subset $ S $ such that $ S \cap K $ is closed for all closed compact $ K $ is  itself closed.
A topological space is called $ KC$ - space  provided every compact sets are closed.
A topological space is called $ US $ - space  provided each convergent sequence has unique
limit.
$ T2⇒ KC ⇒ US ⇒ T1 $
Theorem: Let $ X $ be a $ KC $ space . Then $ X ^ { * } $ ( one point compatification of $ X $ ) is $ KC $ iff $ X $ is a k space.
To show that a $ US $ space is not $ KC$ space. let $ X $  be a non- compact $ T_{2} $ space  which is not a k- space, so according upper theorem $ X^{*} $ is compact $ US $ but not $ KC $ .for example:
A non-compact  $ T_{2} $ space which is not a k - space. Let $ X = N \cup \{ b\} $ , where $ b \in \beta\mathbb{N} - \mathbb{N} $ . If $ S $ is any infinite subset of $ \mathbb{N}$ , neither $ S $ nor $ S \cup \{ b \} $
 is closed in . Hence neither is compact. Thus $ X $ is pseudo - finite ,i.e. all its compact sets are finite. if $ X $ were a k - space , it would , being pseudo- finite, be discrete.

why is it right that :If $ S $ is any infinite subset of $ \mathbb{N}$ , neither $ S $ nor $ S \cup \{ b \} $
   is closed in . Hence neither is compact. Thus $ X $ is pseudo - finite ,i.e. all its compact sets are finite. if $ X $ were a k - space , it would , being pseudo- finite, be disceret." ?

 A: 
why is it right that :If $ S $ is any infinite subset of $ \mathbb{N}$ , neither $ S $ nor $ S \cup \{ b \} $
   is closed in . Hence neither is compact. 

I suppose you meant to say closed in $\beta\mathbb{N}$.
The closure of $S$ in $\beta\mathbb{N}$ consists of all ultrafilters1 containing $S$. If $S$ is infinite, there are at least two such ultrafilters; to see this you can, for example, divide $S$ into two disjoint infinite subsets $S=A\cup B$ and take one free ultrafilter containing $A$ and $S$ and another one containing $B$ and $S$.
You can make basically the same argument using Engelking, Corollary 3.6.2: Every pair of completely separated subsets of a Tychonoff space $X$ has
disjoint closures in $\beta X$.
Two subsets $A$ and $B$ of a topological space $X$ are called completely
separated if there exists a continuous function
$f \colon X\to I$ such that $f(A)=0$ and $f(B)=1$. We say that $f$ separates sets $A$ and $B$.
Clearly, any two disjoint subsets of a discrete space are completely separated.
(Basically all you need for this argument to work is to show that closure of $S$ in $\beta\mathbb N$ contains at least two points from $\beta\mathbb N\setminus\mathbb N$.)

Thus $ X $ is pseudo - finite ,i.e. all its compact sets are finite. if $ X $ were a k - space , it would , being pseudo- finite, be disceret." ?

If we work with a $T_1$-space, all finite subsets are closed. Now we have that for any compact subset $K$ any any subset $V$ the set $V\cap K$ is finite, and thus closed. This shows that every subset $V\subseteq X$ is closed.

1 I am using a construction of $\beta\mathbb N$ using ultrafilters.
The construction of $\beta\mathbb N$ using ultrafilters is briefly described on Wikipedia.
You can find it also in many internet resources and books.
For example, it is described here: 


*

*Chapter I.3 in Hindman N., Strauss D. Algebra in the Stone-Čech compactification

*Chapter II.14 in Todorcevic S. Topics in topology (Lecture Notes in Mathematics 1652).

