the distribution not being defined through a locally integrable function then the support is the complement open set on which $T$ vanishes.
if we consider test function with support not including $[0,1]$ then $T(\phi) = 0$, intuitively the support then must be $[0,1]$ but I'm having difficulties to prove it formally, i.e is to prove that $]-\infty,0[ \cup]1,+\infty[$ is indeed the biggest open set on which $T$ vanishes.
how to prove it ?
If $\phi$ vanishes on $\{1/j: j\in\mathbb N\} \cup\{0\}$ then $T(\phi)=0$. This should help you to guess the support. The interesting point of the example is that -- although the formula only involes values of $\phi$ -- you need derivatives of $\phi$ (also outside the support of $T$) to estimate $T(\phi)$. More precisely, applying the mean value theorem you get $|T(\phi)|\le \zeta(2) \sup\{|\phi'(x)|: x\in[0,1]\}$.