This proof uses least upper-bound property which I am guessing might be equivalent to assuming connectedness of real numbers, since for $\mathbf{Q}$ I would not be able to assume the largest of the $\delta$ to exist.
Assume that the function is continuous, then consider an arbitrary point $x_{0} \in \left[0,1\right]$. Then, for every $\epsilon > 0$, there must exist a $\delta > 0$ such that $0
Let $x_{0} = 0$, then for any $\epsilon$, there exists a $\delta_{0}>0$, hence in the interval $[0,\delta_{0})$, $f(x)=0$. Now we can say that for no $\delta > d_{0}$ is the condition satisfied, because if it is, then we can take the new $\delta$ as $\delta_{0}$ and go on increasing till we get $\delta_{0}=1$.
Now for any interval around $\delta_{0}$, it can be shown that the condition for continuity is not satisfied.