Proof that a continuous map to $S^n$ whose image is a proper subset of $S^n$ is null-homotopic

I am attempting to prove the following:

If $g:X \to S^n$, $n \ge 1$, is a continuous map whose image $g(X)$ is a proper subset of $S^n$, then $g$ is null-homotopic.

Just before this I proved that if $Y$ is contractible, then every continuous map $f:X \to Y$ is null-homotopic. Therefore I'd like to show that any proper subset of $S^n$ is contractible. What is the best way to do this? I wanted to use that $S^n$ is an $n$-manifold to associate it with an open set in $\mathbb{R}^n$, which, if it is convex, is contractible. However, I don't think this works, and it seems like overkill anyway.

What is the best way to show that a proper subset of $S^n$ is contractible?

Thanks.

-
It is not true that any proper subset of $S^n$ is contractible!! What is true is that $S^n - x$ for any point $x$ is contractible, and that is enough. –  Dylan Wilson Feb 13 '13 at 2:28
@Dylan: Of course, I should have realized that... Thanks. –  Alex Petzke Feb 13 '13 at 2:32

Not all proper subsets of $S^n$ are contractible (unless $n = 0$ obviously). Take any two unequal points $x \neq y \in S^n$, then $\{x,y\} \subset S^n$ is not contractible.
What is true is that $S^n \setminus \{ x \} \cong \mathbb{R}^n$ is contractible, so if the map isn't surjective, it factors through a contractible space so it's nullhomotopic.
If $g\neq a\in S^n$ use
$H(x,t )=\dfrac{ tg(x)+(1-t)a}{ \| tg(x)+(1-t)a\|}$