For $n \geq 2$, $\mathbb{R}^n$ and an $n$-dimensional sphere are path-connected Let $n \in \mathbb{N}$, $n ≥ 2$. I want to prove that 
a) $\mathbb{R}^n \setminus \{0\}$ is path-connected. 
b) the sphere $S^{n-1} := \{x \in \mathbb{R}^n: ||x||_2 = 1\}$ is path-connected.
For a), I would need to find a path that connects any two points $x, y \in \mathbb{R}^n$. In case the path that leads directly from $x$ to $y$ doesn't contain $0$, this would be a connection. But if the direct path does contain $0$, I would need to find another way.
Now it's easy to imagine that there are plenty of curves from $x$ to $y$ that don't touch $0$, but what would be a most elegant path where it's easy to see/prove that it's indeed one?

For a) As $n \ge 2$, take a point $z$ which is not on the line $x,y$. Take for your path the union of the two segments $[x,z]$ $[z,y]$.
For b) The application $\varphi$ from $\mathbb{R}^n \setminus \{ 0 \}$ to $S^{n-1}$ defined by $x \to x/\Vert x \Vert_2$ is continuous. If $x,y$ are on the sphere and not on a diameter, the path $\varphi([x,y])$ is continuous and join  $x$ to $y$ on the sphere. If $x$, $y$ are on a diameter, $\varphi([x,z][z,y])$ is a solution.
