$f:\bf S^1 \to \bf R$, 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?

Here is "almost" same solution as of @Anthony Carapetis written in slightly different way:
Assume $f$ is not constant.Since $\bf {S}^1$ is connected and compact so let $f(\bf S^1)=[a,b]. $Suppose pre image of some $y \in (a,b) $ consists only one point,say $x$.Then  $f( \bf {S}^1 / {x})$ is connected.Contradiction!
