We have just started learning about continuity and this is basically the first assignment on this subject so I'm very unsure about how everything works. My attempt of a solution is as follows;
Let $g:\left[0,\frac{1}{2}\right]\ ->\left[0,\frac{1}{2}\right]\ $such that $ g\left(x\right)\ =\ f\left(x\right)\ -\ f\left(x+\frac{1}{2}\right)$ Note that $g$ is continuous, hence by the fixed point theorem there exists a point $C$ in $\left[0,\frac{1}{2}\right]$ such that $g\left(C\right)\ =\ C$.
Therefore there exists a $C$ in $\left[0,\frac{1}{2}\right]$ such that $ g\left(C\right)\ =\ f\left(C\right)\ -\ f\left(C+\frac{1}{2}\right)$.
Now am I missing something? Because it seems just too simple to me.. also I didn't even use the fact the $f(0) = f(1)$ which is suspicious.
Edit: I'm seeing now that i didn't use the theorem correctly, since it just says that $g\left(C\right)\ =\ f\left(C\right)\ -\ f\left(C+\frac{1}{2}\right)= C $. and C is not necessarily $0$..
Your function $g$ may not necessarily be a self-map of $ [0,\frac{1}{2}]$ (and thus you can’t use the fixed point theorem justifiably). Rather it is only guaranteed (as you have noted) that it is continuous on $[0,\frac{1}{2}]$ and satisfies $$g(0)=f(0)-f(\frac{1}{2})\,~\,~\,~ and \,~\,~\,~g(\frac{1}{2})=f(\frac{1}{2})-f(1)\,.$$ Since $f(0)=f(1)$, it follows that $$g(0)=-g(\frac{1}{2})\,.$$ Thus either $g(0)=0$ —— in which case $f(0)=f(\frac{1}{2})$ —— or by the Intermediate Value Theorem there exists $x_0\in (0,\frac{1}{2})$ such that $g(x_0)=0$ — that is $f(x_0)=f(x_0+\frac{1}{2})$.