Show that the one point compactification of $\mathbb{Q}$ is not Hausdorff https://en.wikipedia.org/wiki/Alexandroff_extension
Definition: one point compactification

Let $X$ be any topological space, and let   $ \infty$  be any object
  which is not already an element of $X$. Put $ X^{*}=X\cup \{\infty
 \}$, and topologize $X^*    $ by taking as open sets all the open
  subsets $U$ of $X$ together with all subsets $V$ which contain $\infty
 $  and such that $X\setminus V$ is closed and compact

Show that the one point compactification of $\mathbb{Q}$ which is $\mathbb{Q}^*$ is Not Hausdorff.
What to do? What...
My Attempt:
Suppose to the contrary $\mathbb{Q}^*$ is Hausdorff, then we take two points $x$, $\infty \in \mathbb{Q}^*$, $x \neq \infty$, and produce disjoint open sets $U,V$ such that $x \in U$ and $\infty \in V$
By definition, we know that $\mathbb{Q} \backslash V$ is a closed and compact space such that $x \in U \subseteq \mathbb{Q} \backslash V$
We wish to produce a contradiction such that $\mathbb{Q} \backslash V$ is not closed, or not compact. But we know that $\mathbb{Q} \backslash V$ has to be closed since $V$ is open, therefore we need to show $\mathbb{Q} \backslash V$ is not compact.
Let $\mathcal{U}$ be an open cover of $\mathbb{Q} \backslash V$. Since  $\mathbb{Q} \backslash V$ is claimed to be compact, then $\mathcal{U}$ has a finite subcover $\{U_i|i \in F\}$, $F$ is finite in  $\mathbb{Q} \backslash V$. Then for all $x \in \mathbb{Q}\backslash V$, $\exists i \in F$ s.t. $x \in U_i$ (...Ugh everything seems fine...)
Can someone provide me with some help as to how to go on with this proof? Thanks a bunch.
 A: Given your open sets $U$ and $V$ in $\Bbb Q^*$, you can continue the argument as follows.
$U$ is an ordinary open nbhd of $x$ in $\Bbb Q$, so it contains an open nbhd of $x$ of the form $(a,b)\cap\Bbb Q$ for some irrational $a,b\in\Bbb R$. Let $W=(a,b)\cap\Bbb Q$; clearly $W\cap V=\varnothing$, so in particular $\infty\notin\operatorname{cl}_{\Bbb Q^*}W$. Thus, 
$$\operatorname{cl}_{\Bbb Q^*}W=\operatorname{cl}_{\Bbb Q}W=[a,b]\cap\Bbb Q=W\;.$$
But this is clearly impossible, since $W$ is not compact. 
If you want an explicit example of an open cover of $W$ with no finite subcover, let $\langle a_n:n\in\Bbb N\rangle$ and $\langle b_n:n\in\Bbb N\rangle$ be sequences in $(a,b)$ such that


*

*$\langle a_n:n\in\Bbb N\rangle$ is strictly decreasing and converges to $a$,  

*$\langle b_n:n\in\Bbb N\rangle$ is strictly increasing and converges to $b$, and  

*$a_0<b_0$.


Let $U_n=(a_n,b_n)\cap\Bbb Q$ for $n\in\Bbb N$; then $\{U_n:n\in\Bbb N\}$ is the desired cover.
A: Let $C$ be a compact subset of $\mathbb Q$. Then because the inclusion map $\mathbb Q\to\mathbb R$ is continuous it follows then $C$ is a compact subset of $\mathbb R$, so it is bounded and has all its limit points. 
Let $a,b\in\mathbb R$ with $a<b$ such that $(a,b)\cap\mathbb Q\subseteq C$, then the closure of $C$ in $\mathbb R$ includes $(a,b)$. So by contradiction $C$ has empty interior. Therefore every open set containing $\infty$ is dense.
Finally let $x\in \mathbb Q$ and let $U,V$ be open sets with $x\in U$ and $\infty\in V$, then $V$ is dense so $U\cap V\ne\emptyset$.
