Is $\mathbb{R}$ with the particular point topology and open half line topology compact? Consider $(\mathbb{R}, \tau_{p})$ and $(\mathbb{R}, \tau_l)$ (or sometimes called "ray topology")
Where $\tau_{p} = \{U \subseteq \mathbb{R}, p \in U\}\cup\{\varnothing\}$, and $\tau_l = \{(a, \infty)| a \in \mathbb{R}\}\cup\{\varnothing, \mathbb{R}\}$

Question: Are $(\mathbb{R}, \tau_{p})$ and $(\mathbb{R}, \tau_l)$
  compact?



*

*Let $\mathcal{U}$ be an open cover of $(\mathbb{R}, \tau_{p})$, then it necessarily contains $\{\mathbb{R}\}$, since $p \in \mathbb{R}$. Then we can remove all other open sets, leaving only $\{\mathbb{R}\}$. So every open cover has a finite subcover. 

*Let $\mathcal{U}$ be an open cover of $(\mathbb{R}, \tau_{l})$. Since $\mathbb{R}$ is uncountabe, to produce a finite subcover, we must remove all but finite number of sets in $\mathcal{U}$. Suppose we have removed all but finite number of sets in $\mathcal{U}$.  Since $\mathbb{R}$ has no least element, therefore we can always find $a \in \mathbb{R}$ such that no cover contains it, therefore $(\mathbb{R}, \tau_{l})$ is not compact.
I know my arguments sort of sucks, I would appreciate if someone can check if these are correct and any improvement on my arguments are appreciated!

HINT:


*

*Consider $\mathcal{U}=\{(-n,n): n\in\mathbb{N}\}$. Has it a finite subcover?

*Is already done by you.
