Let $S^n$ denote the unit sphere in the Euclidean space $\Bbb R^{n+1}$, $X$ a topological space, $f,g:X\to S^n$ are both continuous and there doesn't exist $x\in X$ such that $f(x)=-g(x)$, show that $f,g$ are homotopic.
I haven't been able to construct any rigorous proof so far, but I do feel it is geometrically easy. My intuition is, for any $x\in X$, we just need to connect $f(x)$ and $g(x)$ in the following way: suppose $f(x)=(a,h_1),a\in\Bbb R^n, h_1\in \Bbb R$ and $g(x)=(b,h_2)$ similarly, if $h_2\ne -1$, then for $b$ obviously there exists a unique "longitude" passing through it, and for $h_1$ there is also a unique "latitude" through it (of course, here we better interpret $S^n$ as "spanned" from $S^{n-1}$ with the additional $(n+1)$-th dimension being the "height"). Then denoting by $p$ the point where the lattitude and longitude meet, we can simply find a path $\gamma$ on $S^n$ which first connects $f(x)$ to $p$ latitudinally, then $p$ to $g(x)$ longitudinally. Moreover, to determine the explicit homotopy $H(x,t)$, we can split the time domain into $[0,1/2],[1/2,1]$, the first part to be spent on the latitudinal trip and the second the longitudinal one. This entire idea seems quite workable for me.
When I'm trying to formalise this idea into a rigorous proof, it becomes quite messy to deal with the case $h_2=-1$, where we can't guarantee a unique longitude through $g(x)$. Of coure one can always find a longtitude through $g(x)$, but without uniqueness, it would be very troublesome to define a general rule to find one longtitude for each $g(x)$ in a consistent pattern, or in other words, if we allow loose choice of longitude for any special $g(x)$, the resulting "homotopy" might well fall apart due to discontinuity.
So my question is, how to neatly apply the "no antipodal pair" condition to get around this flaw and make the entire idea workable again?