HMMT February, 2015

Solution:
We claim we can choose $\epsilon(f)=\max \left(|f(x_{1})-f(x_{2})|\right) / 2$. Fix $P$ and suppose for the sake of contradiction that for all $z$ with $|z| \leq 1$ it is the case that $|f(z)-P(z)|<\epsilon$. We can write $z=r e^{i \theta}$ so that
$$|f(r)-P(r e^{i \theta})|<\epsilon$$
Let $p$ be a prime larger than $\operatorname{deg}(P)$ and set $\theta=0,2 \pi / p, 4 \pi / p, \ldots$ in the above. Averaging, and using triangle inequality,
$$\left|f(r)-\frac{1}{p} \sum_{k=0}^{p-1} P\left(r e^{2 \pi i k / p}\right)\right|<\epsilon$$
The sum inside the absolute value is a roots of unity filter; since $p>\operatorname{deg}(P)$ it is simply the constant term of $P$ - denote this value by $p_{0}$. Thus for all $r$,
$$|f(r)-p_{0}|<\epsilon$$
Setting $r=x_{1}, x_{2}$ and applying triangle inequality gives the desired contradiction.