From wer!

Translation invariant, hence wlog $f=0$ on ${\bf N}$. If $a<b,f(a)=f(b)=0$ then we can take $d>c>\max(a,b)$ where $c,d$ are integers of the same parity. Hence $f({{a+b}\over 2})=0$. - Doing this for $(a,b)=(0,1),(1,2),\cdots$ we find $f=0$ on ${1\over 2}{\bf N}$. - Doing this for $(a,b)=(0,1/2),(1/2,1),\cdots$ we find $f=0$ on ${1\over 4}{\bf N}$. - Continuing we find that $f=0$ on all dyadic fractions. Since they lie dense and $f$ continuous, we have $f\equiv 0$.

Consider any point any point $c>0$ and let $n\in\mathbb{Z}_{\ge 0}$ such that $n\le f(c)\le n+1$. We will show that $f(c)=f(n)$. Partition the interval $[n,n+1]$ at the midpoint for example $[n,n+1/2]$ and $(n+1/2, n+1]$ and further partition the interval containing $c$ in the same manner and go on doing this. Notice that the sequence of $f$ evaluated at the midpoints of these intervals converge to $f(c)$ and as it's a constant sequence, each term of the sequence is $f(n)$, hence $f$ is a constant function. $\square$