Let $f: [0,2a] \to \mathbb{R}$ be a continuous function with $f(0) = f(2a)$. Show that there exists $x_0 \in [0,a]$ such that $f(x_0) = f(x_0 + a)$.
Anyone have ideas on this? Maybe I can say: Suppose not; then there exist $x_0 \in [0,a]$ s.t. $f(x_0) \neq f(x_0 + a)$. i.e., $f(x_0) - f(x_0 + a) \neq 0$. This makes me think we should use the Intermediate Value Theorem to show that there actually is some zero such that $g(x) = f(x_0) - f(x_0 + a) = 0$.
You seem to be on the right track. We can define \begin{equation*} g(x)=f(x)-f(x+a) \end{equation*} and note that \begin{align*} g(0) &= f(0) - f(a)\\ g(a) &= f(a) - f(2a). \end{align*} Hence \begin{equation*} g(0)+g(a) = f(0)-f(2a) = 0. \end{equation*}
This last equation means that either $g(0)=g(a)=0$ or $g(0)$ and $g(a)$ have opposite signs. In either case, the Intermediate Value Theorem tells us that there is a value $x \in [0,a]$ such that $g(x)=0$, and thus that $f(x)-f(x+a)=0$.
