real analysis - A continuous function $f: [0,1] rightarrow mathbb{R}$ continuous with $f(0) = f(1)$

I came across this question in Stephen Abbott's "Understanding Analysis" 2nd edition.

Let $f: [0,1] \rightarrow \mathbb{R}$ be continuous with $f(0) = f(1)$.

a) Show that there must exist $x,y \in [0,1]$ satisfying $|x-y| = 1/2$ and $f(x)=f(y)$.

b) Show that for each $n \in \mathbb{N}$ there exist $x_n , y_n \in [0,1]$ with $|x_n-y_n|=1/n$ and $f(x_n)=f(y_n)$.

For part a), I thought that I would start by splitting the interval $[0,1]$ into two halves and asserting that if $f(1/2)=c\neq 0$ then there must be some $x \in [0, 1/2]$ and $y \in [1/2,1]$ such that $f(x)=f(y)$ (by the Intermediate Value Theorem) but that did not lead me anywhere.