I have one little problem with one point compactification. Here is example: Let $Z = \alpha \mathbb N = \mathbb N \cup \{\alpha\}$ one point compactification discrete spaces $\mathbb N$ of natural number.
I know that open sets are shapes $\{\alpha\} \cup \mathbb N \backslash K$, where $K$ is compact subset of $\mathbb N$. $\mathbb N$ is discrete space so only finite subsets of $\mathbb N$ are compact.
Question: How to prove that $Z$ is compact?
You need to show that every open cover of the one-point compactification of $\mathbb{N}$ has a finite subcover. The open subsets of the one-point compactification are typically taken to be any open set of the original space (in this case $\mathbb{N}$), or the compactification minus a compact set of the original space (in this case $(\{\alpha\} \cup \mathbb{N}) \setminus C$, where $C$ is a finite set of $\mathbb{N}$).
Let $\mathcal{U}$ be an open cover of $\{\alpha\} \cup \mathbb{N}$. Then there must be some $U_1 \in \mathcal{U}$ such that $\alpha \in U_1$, and looking at the open sets described above, we see that this $U_1$ must exclude only finitely many points of $\{\alpha\} \cup \mathbb{N}$. Can you finish from there?