= Solution
Let $\mathcal A$ be the closure of the <polynomials> in $C(K)$ for the <uniform norm>. It is a closed <algebra>, since products of uniformly convergent sequences converge uniformly on the <compact set> $K$.
Choose a point $W$ with $|W|>\sup_{z\in K}|z|$. The uniformly convergent <geometric series>
$$
\frac1{W-z}=\sum_{j=0}^{\infty}\frac{z^j}{W^{j+1}}
$$
shows that this reciprocal belongs to $\mathcal A$. Join $w$ to $W$ by a <continuous path> in $\mathbb C\setminus K$. Its image is compact and disjoint from $K$, so their distance $d$ is positive. Partition the path into finitely many steps shorter than $d/2$.
Suppose $1/(b-z)\in\mathcal A$ and $|a-b|<d/2$, with $b$ on the path. Then
$$
\frac1{a-z}=\sum_{j=0}^{\infty}\frac{(b-a)^j}{(b-z)^{j+1}}
$$
converges uniformly on $K$, since $|b-z|\geq d$. Every partial sum belongs to $\mathcal A$, by its algebra property, and so does the limit. Working backwards along the partition transfers the pole from $W$ to $w$. Thus $1/(w-z)\in\mathcal A$: \b[a polynomial $P$ exists with]
$$
\boxed{\sup_{z\in K}\left|P(z)-\frac1{w-z}\right|<\epsilon.}
$$
This <pole-moving polynomial approximation> is the elementary mechanism behind the <polynomial Runge theorem>.
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