Showing the Stone-Cech compactification of the naturals minus a point is not paracompact 
I have managed to show 1. and 2. but I am not sure how to go about the last part about showing that the space is not paracompact. How should this be done?
 A: First of all, $\beta \Bbb N$ is separable (as it contains $\Bbb N$ as the countable dense subset). Moreover, $\beta \Bbb N\setminus \Bbb N$ is not first countable at any point (for example, see this answer). 
For the sake of contradiction, assume $\beta\Bbb N\setminus\{p\}$ is paracompact. Since $\mathcal{U}=\{\beta \mathbb{N} \setminus \overline{U} : U \text { is a neighborhood of }  p
\}$ is an open cover of $\beta\Bbb N\setminus\{p\}$ ($\because\beta\Bbb N$ is Hausdorff), it has a locally finite open refinement $\mathcal U'$. After you've proven (1) and (2), you know $\mathcal U'$ is countable and $\displaystyle\bigcup\left\{\text{cl}_{\beta \Bbb N\setminus \{p\}}(U'):U'\in \mathcal U'\right\}=\beta \Bbb N\setminus\{p\}$ (Here $\overline{U'}$ and $\text{cl}_{\beta \Bbb N\setminus \{p\}} (U')$ denotes the closure of $U'$ in $\beta\Bbb N$ and in $\beta\Bbb N\setminus\{p\}$, resp.). 
Then $\displaystyle\bigcup\left\{\overline{U'}:U'\in \mathcal U'\right\}\color{red}{\supseteq}\bigcup\left\{\text{cl}_{\beta \Bbb N\setminus \{p\}}(U'):U'\in \mathcal U'\right\}=\beta\Bbb N\setminus \{p\}$. We note that $p\notin\overline{U'}$ for every $U'\in\mathcal U'$ as we can find some neighborhood $U$ of $p$ such that $U'\subseteq \beta \Bbb N \setminus \overline{U}\Rightarrow U'\cap \overline{U}=\emptyset\Rightarrow U\cap \overline{U'}=\emptyset$. This implies $\displaystyle\bigcup\left\{\overline{U'}:U'\in \mathcal U'\right\}\color{red}{=}\beta \Bbb N\setminus \{p\}$. Taking complements of both sides, we get $\displaystyle\{p\}= \beta \Bbb N\setminus \bigcup\left\{\overline{U'}:U'\in \mathcal U'\right\}=\bigcap \left\{\left(\overline{U'}\right)^c:U'\in \mathcal U'\right\}$. Being a countable intersection of open set, $\{p\}$ has a countable local base (by the statement enclosed in square brackets; a complete proof can be found here), contrary to the fact that $\beta\Bbb N\setminus \Bbb N$ is not first countable at any point.
