Let $f:[0,1] \rightarrow \mathbb{R}$ be a continuous function. If $f(0)=f(1)$ then exists $x\in [0,1]$ s.t. $f(x) = f(x+\frac{1}{2})$ 
Let $f:[0,1] \rightarrow \mathbb{R}$ be a continuous function. If $f(0)=f(1)$ then exists $x\in [0,1]$ s.t. $f(x) = f(x+\frac{1}{2})$

We have that $f\left(\frac{1}{2}\right)=f(0)$ or $f\left(\frac{1}{2}\right)>f(0)$ or $f\left(\frac{1}{2}\right)<f(0)$.
If $f\left(\frac{1}{2}\right)=f(0)$ then $x = 0$ Solves the problem.
if $f\left(\frac{1}{2}\right)<f(0)$ then by the intermediate value theorem 
$$\exists k \in \left[0,\frac{1}{2}\right], f(0)>f(k)>f\left(\frac{1}{2}\right) \implies$$
$$\implies f(1)>f(k)>f\left(\frac{1}{2}\right)\implies \exists t \in \left[\frac{1}{2},1\right] s.t. f(t)=f(k).$$ 
Now how can I show that exists at least one value of k such that $t = k+\frac{1}{2}$? (The solution is similar if $f(\frac{1}{2})>f(0)$)

Let $g(x)=f(x+\frac{1}{2})-f(x)$. It suffices to show some $x\in[0,1]$ vanishes $g(x)$. Note that
$$
g(1/2)g(0)=[f(1)-f(1/2)]\times[f(1/2)-f(0)]=-[f(1/2)-f(0)]^2\leq 0.
$$
If you actually have equality, then you're done: either $x=0$ or $x=\frac{1}{2}$ will make $f(x)=f(x+\frac{1}{2})$. If you have strict inequality, then you're also done thanks to Bolzano's Theorem.
