Proof:

Note that if $T$ is non-singular, it is a proper map and so it extends continuously as a map of $S^n$ sending the point at infinity to itself. Conversely, if $T$ fails to be bijective then there is a sequence of points ${\bf x}_n$ such that $\Vert{\bf x_n}\Vert\rightarrow +\infty$ but $T({\bf x}_n) = 0$ for every $n$. Thus if $T$ were to extend continuously as a map of $S^n$ we would be forced to map the point at infinity namely the north pole to (the point of $S^n$ corresponding to) the origin. On the other hand since $T$ is not the zero map, pick a vector ${\bf u}$ such that $T{\bf u} \neq 0$ and the sequence $m{\bf u}$ converges (as $m\rightarrow \infty$) to the point at infinity on $S^n$. Thus by continuity we would have $\lim T(m{\bf u}) = 0$, as $m\rightarrow 0$. Hence, $m\Vert T{\bf u}\Vert\longrightarrow 0$ which is plainly false since $T{\bf u} \neq 0$.

More important examples are furnished by regarding $S^2$ as the one point compactification of the plane $\mathbb{C}$ and using the field structure on the plane. The proof of the following is an exercise.