I have to show that the map $f(x)=\dfrac{x}{\lVert x\rVert}$ is an open map where $x$ belongs to $\mathbb{R}^{n+1}/{0}$ and the range is $\mathcal{S}^n$ with the subspace topology.
That this map is open seems intuitively clear but it is much more nontrivial for me to show rigorously why it is open...Could anyone please help me?
$\Bbb R^{n+1}\setminus\{0\}$ has the same topology as $\mathcal{S}^n\times(0,\infty)$ endowed with product topology (how?). With some suitable and canonical homeomorphism $h:\Bbb R^{n+1}\setminus\{0\}\to\mathcal{S}^n\times(0,\infty)$, you can show that $f=\pi_1\circ h$, where $\pi_1:\mathcal{S}^n\times(0,\infty)\to\mathcal{S}^n$ is the canonical projection to the first coordinate: $\pi_1(x,y):=x$ for any $x\in\mathcal{S}^n$ and $y\in(0,\infty)$.
Of course $h$ is an open map as homeomorphisms are open maps. The projection map is also an open map. See Projection is an open map.
Composition of two open maps is again open. In our case, $f=\pi_1\circ h$ so $f$ is open.
