Prove that $\mathbb{S}^n$ (with the subspace topology) is the one-point-compactification of $\mathbb{S}^n \setminus \{p\}$ where $p \in \mathbb{S}^n$.
Note that I've already shown on my own that $\mathbb{S}^n \setminus \{p\}$ is locally compact.
Proof: Consider $(\mathbb{S}^n, \kappa)$ where $\kappa$ is the subspace topology on $\mathbb{S}^n$. Now consider the one-point compactification $\mathbb{S}^n \setminus \{p\} \cup \{p\} = \mathbb{S}^n$ equipped with the topology $\mathcal{T} = \{\text{open subsets of } \ \mathbb{S}^n \setminus \{p\}\} \cup \{ U \subseteq \mathbb{S}^n \ | \ \mathbb{S}^n \setminus U \text{ is a compact subset of } \mathbb{S}^n \setminus \{p\}\}$. I claim that $(\mathbb{S}^n, \kappa)$ is homeomorphic to $(\mathbb{S}^n, \mathcal{T})$
Since $\{p\}$ is closed in $\mathbb{R}^{n+1}$ and $p \in \mathbb{S}^n$ and $\{p\} = \mathbb{S}^n \cap \{p\}$ it follows that $\{p\}$ is closed in $\mathbb{S}^n$. Thus $\mathbb{S}^n \setminus \{p\}$ is open in $\mathbb{S}^n$. Thus the open subsets of $\mathbb{S}^n \setminus \{p\}$ are all open in $\mathbb{S}^n$ with the subspace topology.
We now claim that $$\{\text{neighborhoods of $p$ in $(\mathbb{S}^n, \kappa)$}\} = \{ U \subseteq \mathbb{S}^n \ | \ \mathbb{S}^n \setminus U \text{ is a compact subset of } \mathbb{S}^n \setminus \{p\}\}$$
Let $U$ be a neighborhood of $p$ in $(\mathbb{S}^n, \kappa)$ then $\mathbb{S}^n \setminus U$ is closed in $(\mathbb{S}^n, \kappa)$ and since $(\mathbb{S}^n, \kappa)$ is compact it collows that $\mathbb{S}^n \setminus U$ is compact and moreover $\mathbb{S}^n \setminus U \subseteq \mathbb{S}^n \setminus \{p\}$, thus $\{\text{neighborhoods of $p$ in $(\mathbb{S}^n, \kappa)$}\} \subseteq \{ U \subseteq \mathbb{S}^n \ | \ \mathbb{S}^n \setminus U \text{ is a compact subset of } \mathbb{S}^n \setminus \{p\}\}$.
Conversely let $V \subseteq \mathbb{S}^n$ be such that $\mathbb{S}^n \setminus B$ is a compact subset of $\mathbb{S}^n \setminus \{p\}$ we show that $V$ is a neighborhood of $p$ in $(\mathbb{S}^n, \kappa)$. Since $\mathbb{S}^n \setminus V$ is compact and $\mathbb{S}^n \setminus V \subseteq \mathbb{R}^{n+1}$ by the Heine-Borel theorem we have that $\mathbb{S}^n \setminus V$ is closed in $\mathbb{R}^{n+1}$ and thus $\mathbb{S}^n \cap \left(\mathbb{S}^n \setminus V\right)= \mathbb{S}^n \setminus V$ is closed in $(\mathbb{S}^n, \kappa)$. Furthermore $\mathbb{S}^n \setminus V \subseteq \mathbb{S}^n \setminus\{p\} \implies p \in V$. Putting all this together we see that $V$ is a neighborhood of $p$ in $(\mathbb{S}^n, \kappa)$ and thus we've shown that $\{\text{neighborhoods of $p$ in $(\mathbb{S}^n, \kappa)$}\} \supseteq \{ U \subseteq \mathbb{S}^n \ | \ \mathbb{S}^n \setminus U \text{ is a compact subset of } \mathbb{S}^n \setminus \{p\}\}$ and proven our claim above.
Thus $$\mathcal{T} = \{\text{open in $\mathbb{S}^n$ not containing $p$}\} \cup \{\text{open sets in $\mathbb{S}^n$ containing $p$}\}$$
and it's clear to see that $\mathcal{T} = \kappa$. Thus $(\mathbb{S}^n, \kappa)$ is indeed homeomorphic (through he identity map) to $(\mathbb{S}^n, \mathcal{T})$ and thus $\mathbb{S}^n$ (with the subspace topology) is the one-point-compactification of $\mathbb{S}^n \setminus \{p\}$ where $p \in \mathbb{S}^n$. $\square$
Is the above proof correct and satisfactory? I know I've skipped a few details, as including those details would make the proof quite lengthy.
An easier proof (that I only realized after writing the above proof) would be to use the following theorem:
Theorem: If $X$ is a locally compact Hausdorff space and $Y$ is compact Hausdorff such that for some point $p \in Y$, $X$ is homeomorphic to $Y\setminus \{p\}$, then $Y$ is homeomorphic to the one-point compactification of $X$.
and then the proof becomes a lot simpler I believe.
