Undergrad algebraic topology proof

This is about Borsuk–Ulam theorem: 

https://en.wikipedia.org/wiki/Borsuk%E2%80%93Ulam_theorem

i.e. for any continuous function $f: \mathbb{S^n}\rightarrow \mathbb{R}^n$ , there exists $x$ so that $f(x)=f(-x)$. For $n=1$ this could be proved by the intermediate value theorem. Indeed, define $f: \mathbb{S}\rightarrow \mathbb{R}$ by
\[g(x)=f(x)-f(-x).\]
If $g \equiv 0$, we are done. If not, then there exists $x\in S$ such that $g(x) g(-x)=-(f(x)-f(-x))^2$. Since $g$ changes sign on the unit circle $S$, there must be a point $x_0$ such that 
\[0=g(x_0)=f(x_0)-f(-x_0)=0 \Rightarrow f(x_0)=f(-x_0).\]
(b) For any line $L: ax+by=c$, by scaling we may assume $a^2+b^2+c^2=1$, i.e. $(a,b,c)\in S^2$. Define

\[U_r=\{(x,y)\in U:  ax+by>c \} ,  U_l=\{(x,y)\in U: ax+by<c \}. \]
Similarly we can define $V_r$ and $V_l$. Define $f: S^2 \rightarrow R^2$ by 
\[f(a,b,c)=(Area(U_r), Area (V_r)).\]
Then we can check that 
\[f(-a,-b,-c)=(Area(U_l), Area (V_l)).\]
By Borsuk–Ulam theorem there exists $(a,b.c)\in S^2$ such that $f(a,b,c)=f(-a-,b,-c)$. So for the line $ax+by=c$ we have
\[(Area(U_l), Area (V_l))=(Area(U_r), Area (V_r)).\]