I'd like to understand the first proof given to me of the fact that the one point compactification of $\mathbb{R}^{n}$ is homeomorphic to $\mathbb{S}^{n}$.
The proof goes as follows : there is an initial remark about $i: \mathbb{R}^{n} \longrightarrow \mathbb{S}^{n}$ being an open embedding (with the identification of $\mathbb{R}^{n}$ as $\mathbb{S}^{n}-\left\lbrace x_{0}\right\rbrace$) and then it states that we only have to proof that the euclidean topology of $\mathbb{S}^{n}$ coincides with the Alexandrov topology on the compactification of $\mathbb{R}^{n}$.
I don't understand how checking the open subset's conditions is sufficient to deduces the homomorphism. Are we using some uniqueness of the Alexandrov topology ?
However I know there is a much simpler way, which is to proof in general that if a topological space $X$ is compact and Hausdorff then it is homeomorphic to the one point compactification of $X$ minus a point, but I'm interested in understanding this one.
Any help or hint would be appreciated.