$f : I=[0,1] \rightarrow \mathbb{R}$ Prove that the lower sum is $0$

Question

Given a function $f : I=[0,1] \rightarrow \mathbb{R}$ defined by
$f(x) = 1$, if $x\in \mathbb{Q}$
$f(x) =0$, if $x\in I \setminus \mathbb{Q}$.
Prove that for every partition $V$ of $I$ the following holds: \begin{align} \underline{S} (f,V) = 0 \end{align} where $\underline{S}(f,V)$ is the lower sum.
If this is the case, I think I need to prove that there will always be an irrational number in every subinterval and this will imply that $\underline{S} (f,V) = 0$.
Is this the correct way to prove that $\underline{S} (f,V) = 0$?