Let $\overline{\Bbb C^2}$ be the one point compactification of $\Bbb C^2$, that is $\overline{\Bbb C^2}=\Bbb C^2\cup\{\infty_2\}$.
It is well known that stereographical projection allows to identify $\overline{\Bbb C}$ (the Riemann sphere) with the unitary ball $\Bbb S^2$ in $\Bbb R^3$: the identifications makes $\infty$ correspond with the north pole $N$ and all the other points $z\in\overline{\Bbb C}\setminus\{N\}$, with the intersection $r_z\cap\Bbb S\setminus\{N\}$, where $r_z$ is the line in $\Bbb R^3$ joining $N$ and $(x,y,0)$ where $z=x+iy$.
Mimic this procedure: take $(z,w)\in\Bbb C^2$, where $z=x_1+ix_2$ and $w=x_3+ix_4$, so we can identify $\Bbb C^2$ as the hyperplane $\{(x_1,x_2,x_3,x_4,0)\}$.
Let the sphere be $\Bbb S^4=\{(x_1,\dots,x_5)\;:\;x_1^2+\cdots+x_5^2=1\}\subset\Bbb R^5$.
Then the argument as above works straightforwardly: identify $\infty_2$ with $N=(0,0,0,0,1)$ and given a point $P=(z,w)\in\overline{\Bbb C^2}\setminus\{\infty_2\}$, we consider the line thru $N$ and $P$, that is $r_P=\{N+t(P-N),t\in\Bbb R\}$, we intersect it with $\Bbb S^4$ and we get two solutions: $t=0$ and $t=2/(x_1^2+\cdots+x_5^2+1)$. So it seems the identification
$$ \overline{\Bbb C^2}\longleftrightarrow\Bbb S^4\subset\Bbb R^5 $$
works fine. Do you confirm that? Or is there something I miss?
Yes, it works, but it has nothing to do with $\mathbb C^2$. In fact, for each $n$ the stereographic projection identifies the one point compactification of $\mathbb R^n$ with $S^n$. Your result follows for $n = 4$.
The stereographic projection homeomorphism is given by $$p: S^n \setminus \{N\} \to \mathbb R^n, p(x) = \left(\dfrac{x_1}{1-x_{n+1}},\ldots,\dfrac{x_n}{1-x_{n+1}}\right) .$$
See
Formula for the intersection of a sphere with regards to stereographic projection
One point compactification of $\mathbb{R}^n$ proof