A continuous map $f:S^2\rightarrow S^2$ such that $f(x)\neq f(-x)$ for all $x$ is surjective 
For all $x\in S^2$, let $-x$ denote its antipode. Let $f:S^2\rightarrow S^2$ be a continuous map such that $f(x)\neq f(-x)$ for all $x\in S^2$. Show that $f$ must be surjective.

I'm working through an old practice exam, and I've been stuck on this question for hours.
I think this has something to do with the Borsuk Ulam Theorem, or maybe Brouwer degree but I am not sure how to construct a function $S^2\rightarrow \mathbb{R}$ that makes $f$ not being surjective a contradiction, or that contradicts anything about degrees.
I've found a somewhat similar question here, but not sure how to adapt it to this problem.
I know that if $f$ is not surjective, $\exists y\in S^2$ such that $y\notin f(S^2)$, so I can construct a well-defined map
$$g:S^2\rightarrow S^2,\hspace{2cm} g(x)=\frac{f(x)-y}{|f(x)-y|}$$
that's homotopic to $f$. Not sure how that helps.
I also thought of the map
$$h:S^2\rightarrow S^2,\hspace{2cm}h(x)=\frac{f(x)-f(-x)}{|f(x)-f(-x)|}$$
which we can well-define, and is homotopic to both $f$ and $f\circ A$ where $A:x\rightarrow -x$.
I also know that if $f$ is not surjective then its degree is $0$, and that the degree of homotopic maps is equal, so the degree of $g$, $h$ and $f\circ A$ is zero, but not sure if there's anything wrong with that either.
The most elementary answer possible would be most appreciated.

It sounds like you have the right idea! Particularly in thinking about the Borsuk-Ulam theorem. Here's a hint:
Remember that $S^2 \cong \mathbb{R}^2 \cup \{ \infty \}$ (this is stereographic projection).
So (towards a contradiction) say we're given a map $f : S^2 \to S^2$ isn't surjective. Well without loss of generality we can say it misses $\infty$, so that composing with stereographic projection gets us a map $f' : S^2 \to \mathbb{R}^2$.
Now, since $\forall x . f(x) \neq f(-x)$, we see that the same must be true for $f'$...
Let me know if you need more than this, but I'll leave it here so it's still technically a hint :P Do you see how to get a contradiction from here?

I hope this helps ^_^
