= Solution
The prescribed-pole form of <Runge theorem> is this: let $K\subset\mathbb C$ be compact, and let $E\subset\widehat{\mathbb C}\setminus K$ meet every <connected component> of the complement in the <Riemann sphere>. If $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon>0$ there is a <rational function> with all its poles in $E$ whose uniform distance from $f$ on $K$ is less than $\epsilon$. A polynomial is regarded as having its only possible pole at infinity. The empty compact set is trivial, so assume $K\ne\varnothing$.
First suppose $\infty\in E$. Let $A\subset C(K)$ be the <uniform rational approximation algebra with prescribed poles>, the uniform closure of <rational functions> whose finite poles lie in $E$. It is a commutative unital <Banach algebra>, and contains $u(z)=z$. Set
$$
S=\{\lambda\in\mathbb C\setminus K:(u-\lambda1)^{-1}\in A\},
$$
where the inverse in this definition is the pointwise <continuous function> on $K$. The set $S$ is relatively open: if the inverse at $\lambda$ belongs to $A$, the <Neumann series> gives inverses at nearby points. It is relatively closed: when $\lambda_n\to\lambda\notin K$, the corresponding scalar functions converge uniformly on $K$, and $A$ is closed.
Each bounded complementary component meets $E$ in a finite point $a$, and $(u-a1)^{-1}$ is an allowed <rational function>. The unbounded component meets $S$ because for $|\lambda|>\|u\|$ the geometric series for $(u-\lambda1)^{-1}$ belongs to $A$. Being both open and closed in the complement, $S$ therefore contains every component, so it is the entire complement. Conversely, evaluation at any $z\in K$ prevents $u-z1$ from being invertible. Hence
$$
\boxed{\sigma_A(u)=K.}
$$
Apply the <holomorphic functional calculus> in $A$ to $f$ near $K$. For every $z\in K$, the evaluation <character of an algebra> and the contour formula from (b) give $f(u)(z)=f(z)$. Thus $f|_K\in A$. By the definition of this uniform closure, the required rational approximations exist.
If $\infty\notin E$, choose a finite point $a\in E$ in the component containing infinity. The <Möbius transformation> $w=(z-a)^{-1}$ sends $K$ to a compact subset $K'$ of the plane and sends $E$ to a set $E'$ containing infinity. It preserves complementary components, so $E'$ meets every component of the complement of $K'$. The transformed function $F(w)=f(a+1/w)$ is holomorphic near $K'$, since $0\notin K'$. The case already proved approximates $F$ by <rational functions> with poles in $E'$. Pulling them back gives rational approximations to $f$ on $K$ with poles only in $E$: a pole at infinity in the $w$-plane becomes a pole at $a$, and every other pole has its prescribed preimage. In particular the pullbacks do not acquire a pole at the original infinity, since $0\notin E'$.
This proves the full prescribed-pole theorem, including the case where a pole at infinity is forbidden. If $\mathbb C\setminus K$ is connected, take $E=\{\infty\}$ to obtain the <polynomial Runge theorem> as a corollary.
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