Proof that the map is an open map. We were asked to prove that the map $f\ :\mathbb{R}^{n+1}/{0}\longrightarrow S^{n}$ given by $f(x)\ =\ x/||x||$ is an open map.
My approach :- it is enough to show that for any $x\in \mathbb{R}^{n+1}$ and the open ball $B(x,R)$, $f(B(x,R))$ is open in $S^n$, i.e. $X\ =\ S^n/f(B(x,R))$ is closed.
So, let $y\in cl(X)$ where $cl(X)$ denotes the closure of $X$. Then there exist a sequence $(y_n)$ in $X$ such that $(y_n)\rightarrow y$. We need to show that $y\in X$ or $y\ =\ z/||z||$ such that $||x-z|| \geq R$. Now $(y_n)$ is in $X$ so $\forall \ n$ $y_n\ =\ z_n/||z_n||$ such that $||x-z_n||\geq R$. Since $(y_n)\rightarrow y$ so $(z_n)$ must converge to $z$ and by the continuity of the norm function $||z_n||\rightarrow ||z||$. Again, by the continuity of the norm function $||x-z||\geq R$ and hence $y\in X$ and so $X$ is closed.
My question is that what is wrong with the above proof ? What would be an alternate argument for showing that $f$ is an open map ?
Thanks.   

The convergence of the $y_n$ doesn't guarantee convergence of $z_n$, e.g. $z_n=n$ doesn't converge, but the corresponding $y_n$ sequence is constant $1$.

Edit: As an alternative proof you can show the following: Suppose $B(x,\delta) \subseteq U$, then $B(f(x), {\delta \over \lVert x \rVert})\cap S^n \subseteq f(U)$.
