$f:\bf S^1 \to \bf R$, there exist uncountably many pairs of distinct points $x$ and $y$ in $\bf S^1$ such that $f(x)=f(y)$? (NBHM-2010) Let $\bf S^1$ denote the unit circle in the plane $\bf R^2$. True/False ?

For every continuous function $f:\bf S^1 \to \bf R$, there exist uncountably many pairs of distinct points $x$ and $y$ in $\bf S^1$ such that $f(x)=f(y)$

Borsuk-Ulam or by taking the function $g(x)=f(x)-f(-x)$, IVT implies that there exist $x$ such that $f(x)=f(-x)$. But I'm unable to show the existence of uncountably many pairs. I think the fact $RP^1 \cong \bf S^1$ may be helpful. Any ideas?

If $f$ is constant then this is obviously true. 
Otherwise, choose points $a,b \in \Bbb S^1$ such that $f(a)\ne f(b)$ and "split the circle" at $a$, so that $f$ becomes a continuous map $[a,a+2\pi] \to \Bbb R$ with $f(a)=f(a+2\pi)\ne f(b)$ for $b\in(a,a+2 \pi)$. Then for any value $M$ strictly between $f(a)$ and $f(b)$, the intermediate value theorem gives us the existence of $x \in (a,b)$ and $y \in (b,a+ 2\pi)$ such that $f(x)=f(y)=M$. Since $f(a)\ne f(b)$, there are uncountably many such values $M$ and thus such pairs, so the proposition is true.
