Problem 1251

If $f(x) = ax^2 + bx + c$ and $v = -b/2a$ is the abscissa of parabola's vertex, then, taking $I = [v - 1/2, v + 1/2]$, the interval $f(I)$ has length $|a|/4$, hence $|a| \ge 4$.

Now, taking an interval $J$ of length 2, one can find $x, y \in J$ such that $x - y = 1$ and $v \notin (y, x)$.
We have $|f(x) - f(y)| = |a(x - y)(x + y + \frac{b}{a})| \ge 4|x + y - 2v| \ge 4$, therefore $f(J)$ contains two points at least 4 units apart, whence the conclusion.