Let $f(x)=p(x-q)^2+r$, for real constants $p, q, r$. Then we know that the image of $[q-\frac{1}{2}, q+\frac{1}{2}]$ has length at least one. This image is $[r, r+\frac{p}{4}]$, or $[r+\frac{p}{4}, r]$, so $|\frac{p}{4}| \geq 1$, or $p \geq 4$. Any interval of length $2$ must contain either an interval of length $1$ where every number is at least $q$, or an interval of length $1$ where every number is at most $q$ - that is, $[q+s, q+s+1]$ or $[q-s-1, q-s]$ for some $s \geq 0$. Then the length of the image of the interval of length $2$ is at least the length of the image of $[q+s, q+s+1]$ or $[q-s-1, q-s]$, which is at least $|f(q \pm (s+1)) - f(q \pm s)| = |(p(s+1)^2 + r) - (ps^2 + r)| = |p(2s+1)| = |p||2s+1| \geq |p| \geq 4$.