Let $\omega\mathbb{Q}$ denote the one-point compactification of $\mathbb{Q}$, where $\omega$ is a point not in $\mathbb{Q}$, and define a topology $\tau$ on $\omega\mathbb{Q}$ by $$ \tau=\{U\subseteq \mathbb{Q}\ |\ U\text{ is open in }\mathbb{Q}\}\cup \{\omega\mathbb{Q}\backslash K\ |\ K\text{ is a compact subset of }\mathbb{Q}\}.$$
Is it true that any compact subset of $\omega\mathbb{Q}$ is then closed?
If $C\subseteq \omega\mathbb{Q}$ is compact and $\omega\notin C$, I think that $C$ will then be compact in $\mathbb{Q}$, so the complement of $C$ is an element of $\tau$, hence $C$ is closed. However, if $\omega\in C$, I'm not sure what to do. I think I'd need to show the complement of $C$ is an open set in $\mathbb{Q}$, which I believe to be true, but I can't prove this explicitely.
Let’s continue. If $\omega\in C$ and $\Bbb Q\setminus C$ is not open in $\Bbb Q$ then $\Bbb Q\cap C$ is not closed in $\Bbb Q$, so there exists a point $x\in \Bbb Q\setminus C$ and a sequence $\{x_n\}$ of points of $\Bbb Q\cap C$ converging to the point $x$. A set $K=\{x\}\cup \{x_n\}$ is a compact subset of $\Bbb Q$. So a family $\{\omega\Bbb Q\setminus K\}\cup\{(\Bbb Q\setminus K)\cup \{x_n\} \} $ is an open cover of the compact $C$, containing no finite subcover, a contradiction.