Mapping unit circle in $\mathbb{R}^2$ to a number in $\mathbb{R}$ is not one-to-one Let $C:=\{\vec{x}\in \mathbb{R}^2:\lVert \vec{x} \rVert=1\}$. Show that a continuous function $f:C\to S\subset\mathbb{R}$ is not one-to-one.

My approach:

Let $f(\vec{x})\in [a,b]$. Let $g:\mathbb{R}\to\mathbb{R}$ be defined as $g(x)=x^2$. Then $g$ is continuous. $(g\circ f)(\vec{x})=[f(\vec{x})]^2\in [a^2, b^2]$. By the Intermediate Value Theorem, $\forall z\in[a^2,b^2],\exists c\in [a,b]$ such that $g(c)=z\implies z = [f(\vec{x_0})]^2=c^2\implies \pm f(\vec{x_0})=c\implies \pm \vec{x}_0 \in f^{-1}(c)\cap C$.
Of course this proof is flawed. I'm not even utilizing the fact that $\vec{x_0}\in C$, so this can apply to any function... Would appreciate some help with this proof.
 A: The result holds in more generality,called the Borsuk-Ulam Theorem
A: Here is a basic argument. With $f$ being a continuous function from a compact connected set $C$ to $\Bbb R$, its image is a finite closed interval $[a,b]$ for some $a,b\in\Bbb R$ with $a\leq b$. Since a constant function on $C$ is not injective, the interval is not reduced to a point: $a\neq b$. There are distinct points $x_a,x_b\in C$ with $f(x_a)=a$ and $f(x_b)=b$. The points $x_a,x_b$, like any pair of distinct points on $C$, delimit two different arcs of the circle (with only those two points as their intersection). The intermediate value theorem can be applied to each arc separately, producing for each a point$~x$ in the interior of the arc with $f(x)=\frac{a+b}2$. This gives two distinct points of $C$ with the same image under$~f$, proving that $f$ is not injective. 
A: Suppose $f$ is injective, and without loss of generality, assume $0 \in f(S^1)^\circ$. The set $A=S^1 \setminus f^{-1}(\{0\})$ is connected, but $f(A)$ is not, hence $f$ is not continuous.
Another approach:
Let $x_0 = \min f(S^1), x_1 = \max f(S^1)$. Let $f(s_0) = x_0, f(s_1) = x_1$.
There are two paths (that are disjoint except for the start and end) joining $s_0, s_1$, call them $\gamma_0, \gamma_1$. Then
$f(\gamma_0((0,1))) = (x_0,x_1), f(\gamma_1((0,1))) = (x_0,x_1)$, hence
for any $x^* \in (x_0,x_1)$ there are points $t_0, t_1$ such that
$f(\gamma_0(t_0)) = x^* = f(\gamma_1(t_1))$. Since $\gamma_0(t_0) \neq \gamma_1(t_1)$ this contradicts injectivity.
