= Solution
Let
$$
S=\{x+iy:-1\leq x,y\leq1\},
$$
and choose $0<\eta<\delta$. Write $C_\eta$ for the positively oriented boundary of the square with $|x|,|y|\leq1+\eta$. The <Cauchy integral formula> gives, for every $z\in S$,
$$
f(z)=\frac1{2\pi i}\int_{C_\eta}\frac{f(\zeta)}{\zeta-z}\,d\zeta.
$$
Because $C_\eta$ and $S$ are compact and separated by a positive distance, Riemann sums for this contour integral converge uniformly in $z\in S$. Thus, for any $\varepsilon>0$, there are points $\zeta_j\in C_\eta$ and constants $c_j$ such that
$$
\sup_{z\in S}\left|f(z)-\sum_{j=1}^m\frac{c_j}{z-\zeta_j}\right|<\frac\varepsilon2.
$$
It remains to approximate each resolvent by polynomials on $S$. Let $\Lambda$ be the set of $\lambda\notin S$ for which $(z-\lambda)^{-1}$ is polynomially uniformly approximable on $S$. If $|\lambda|>\sup_{z\in S}|z|$, the <geometric series>
$$
\frac1{z-\lambda}
=-\frac1\lambda\sum_{n=0}^{\infty}\left(\frac z\lambda\right)^n
$$
shows that $\lambda\in\Lambda$. Moreover, the <resolvent propagation across a complementary component> argument shows that $\Lambda$ is both open and closed relative to $\mathbb C\setminus S$. The complement of a square is <connected>, and $\Lambda$ is nonempty, so $\Lambda=\mathbb C\setminus S$.
In particular, every $\zeta_j$ lies in $\Lambda$. Approximate the finitely many resolvents accurately enough that their weighted sum differs by less than $\varepsilon/2$. Combining this with the Riemann-sum estimate produces a polynomial $p$ satisfying
$$
\sup_{z\in K}|f(z)-p(z)|
\leq\sup_{z\in S}|f(z)-p(z)|
<\varepsilon.
$$
This proves the claimed <elementary polynomial approximation on a square>.
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