This might be long but it's an explicit way to prove this . When f has a fixed point there is nothing to show . Consider the case when f does not have a fixed point . In this case f is homotopic to the antipodal map . Now we consider two cases . First when n is odd . When n is odd the antipodal map is homotopic to the identity map. So we need to worry about the case when n is even i.e. n =2m (say). Now consider the following homotopy
$H^{'}(x,t) : S^{2m} \times [0,\pi] \rightarrow S^{2m}$
$((x_{1},x_{2},...,x_{2m+1}),t) \mapsto (x_{1}Cost-x_{2}Sint,x_{2}Cost+x_{1}Sint,...,x_{2m-1}Cost-x_{2m}Sint,x_{2m}Cost+x_{2m-1}Sint , -x_{2m+1})$
So see that this map is well defined i.e. the image of a point lies in $S^{2m}$ . See that at $t = \pi$ the map is the antipodal map and at t = 0 the map is g ($\textit{say}$) i.e. $g : S^{2m} \rightarrow S^{2m}$ where
$ g(x_{1},x_{2},...,x_{2m+1}) = (x_{1},x_{2},...,-x_{2m+1}) $
So f is homotopic to the antipodal map and further when n is even the antipodal map is homotopic to the map g . So $ f \sim g $ . But see that g fixes all the points $ (x_{1},x_{2},...,x_{2m+1}) $ on $S^{2m}$ when $ x_{2m+1} = 0$ . So f is homotopic to a map that fixes point .