Is my construction of one-point compactification of $\mathbb{R}$ correct? (+ Clarifying questions) Could you please check if this construction makes sense and answer to questions to the parts in bold?
My construction:
The construction can be given explicitly as an inverse stereographic projection. Consider the map
$s: \mathbb{R} \rightarrow S^1$ given by
$x \rightarrow (\frac{1-x^2}{1 + x^2},\frac{2x}{1 + x^2})$
Then $s$ is a homeomorphism between $\mathbb{R}$ and $S^1 \setminus \{(-1,0)\}$. So, since $S^1 \setminus \{(-1,0)\}$ is dense in $S^1$ and $S^1 \setminus\left(S^1 \setminus\{(-1,0)\}\right)$ consists of a single point, $S^1$ is the one-point compactification of $\mathbb{R}$.
Question 1: Why is s a homeomorphism? (How to prove that?) I take this as a known fact from math. analysis, but I am not sure.
Question 2: How do we know that $S^1 \setminus \{(-1,0)\}$ is dense in $S^1$?

Stereographic projection from $p = (-1,0)$ is given by
$$\sigma : S^1 \setminus \{p\} \to \mathbb R, \sigma(x,y) = \frac{y}{1+x}.$$
Your map $s$ has the property $p \notin s(\mathbb R)$ (because $s(x)$ has second coordinate $0$ only for $x = 0$, and $s(0) = (1,0)$). Thus we can regard $s$ as a map $s : \mathbb R \to  S^1 \setminus \{p\}$. Noting $x^2 + y^2 = 1$ for $(x,y) \in S^1$ we get
$$s(\sigma(x,y)) = s\left(\frac{y}{1+x}\right) = \left(\frac{1- (\frac{y}{1+x})^2}{1+ (\frac{y}{1+x})^2}, \frac{2\frac{y}{1+x}}{1+ (\frac{y}{1+x})^2}\right) =  \left(\frac{(1+x)^2 -y^2}{(1+x)^2 +y^2}, \frac{2(1+x)y}{(1+x)^2 +y^2}\right) \\=\left(\frac{(1+2x +x^2 -y^2}{1+2x +x^2 +y^2}, \frac{2(1+x)y}{1+2x +x^2 +y^2}\right) = \left(\frac{(2x +2x^2}{2+2x}, \frac{2(1+x)y}{2+2x}\right) =(x,y), $$
$$\sigma(s(x)) = \sigma\left(\frac{1-x^2}{1+x^2},\frac{2x}{1+x^2} \right) = \frac{\frac{2x}{1+x^2}}{1+\frac{1-x^2}{1+x^2}} = \frac{2x}{1+x^2 + 1 -x^2} = x .$$
This shows that $s$ and $\sigma$ are inverse to each other, i.e. both maps are homeomorphisms.
To see that $S^1 \setminus \{p\}$ is dense in $S^1$, observe that the points $p_n = \left(-\sqrt{1 -\frac{1}{n^2}},\frac 1 n \right)$ are in $ S^1 \setminus \{p\}$ and $p_n \to p$.
