$f : S^1 \to\mathbb R$ is continuous then $f(x)=f(-x)$ for some $x\in S^1$
Define $g:[0,1]\to \mathbb{R}$ as $g(t):=f(e^{2\pi it})-f(-e^{2\pi it})$.
Clearly $g$ is also continuous. We claim that $g$ is zero at some point. If that is true, say $g(t_0)=0$, then it means $f(e^{2\pi it_0})=f(-e^{2\pi it_0})$, hence we get what we wanted.
Suppose $g(t)\neq 0\; \forall\; t\in S^1$, in particular $g(0)\neq 0$. Observe that $g(0)=f(1)-f(-1)$ and $g(\frac{1}{2})=f(-1)-f(1)=-g(0)$. Hence if $g(0)>0$, then $g(\frac{1}{2})<0$, and similarly, if $g(0)<0$, then $g(\frac{1}{2})>0$. Thus by the intermediate value theorem, there must be a point say $t_0\in (0,\frac{1}{2})$ such that $g(t_0)=0$, which gives a contradiction.
Observe that in the same way you can obtain not just a $t_0\in (0,\frac{1}{2})$ such that $g(t_0)=0$, but another point $t_1\in (\frac{1}{2},1)$ such that $g(t_1)=0$, because $g(1)=g(0)$.
Note: Instead of defining $g$, you may define, $h:S^1\to \mathbb{R}$, as $h(x)=f(x)-f(-x)$, and follow the same steps as above. The intermediate value theorem can be applied to the function $h$ also, because $S^1$ is a path-connected space.
You can parameterize $S^1$ in the usual way, so that $f$ is a function from $[0,2\pi]$ to $\mathbb{R}$ with $f(0)=f(2\pi)$.
Then define $g(x):[0,\pi]\to\mathbb{R}$, $g(x) = f(x+\pi)-f(x)$.
- How does $g(0)$ relate to $g(\pi)$?
- What can you conclude from the intermediate value theorem?
Hint: Let $\alpha:[0,1]\to S^1$ parametrize the unit circle, and take $g(z)=f(z)-f(-z)$. Apply the Intermediate Value Theorem to $g\circ\alpha$.
Second hint: If $\alpha(a)=1$ and $\alpha(b)=-1$, analyze $g\circ\alpha(a)$ and $g\circ\alpha(b)$.