= Solution
The prescribed-pole form of <Runge theorem> is as follows. Let $K\subset\mathbb C$ be <compact>, let $f$ be <holomorphic> on a <neighborhood> of $K$, and let $S\subset\widehat{\mathbb C}\setminus K$ meet every <connected component> of that complement. Then, for every $\epsilon>0$, there is a <rational function> $R$, all of whose <poles> belong to $S$, with $\sup_K|f-R|<\epsilon$. A <pole> at infinity means a <polynomial> part is allowed. In particular, if the complement is <connected>, choose $S=\{\infty\}$ and obtain the <polynomial Runge theorem>.
First approximate using <poles> anywhere off $K$. Choose a bounded polygonal <neighborhood> $V$ of $K$ whose closure lies in the <neighborhood> on which $f$ is <holomorphic>, and whose boundary is disjoint from $K$. Such a $V$ comes from a sufficiently fine finite grid covering a small closed <neighborhood> of $K$. Orient the outer boundary positively and boundaries of holes negatively. The <Cauchy integral formula> gives
$$
f(z)=\frac1{2\pi i}\int_{\partial V}\frac{f(\zeta)}{\zeta-z}\,d\zeta,\qquad z\in K.
$$
Because $\partial V$ and $K$ have positive distance, the integrand is <uniformly continuous> on their product, separately along each boundary edge. Its <Riemann sums> converge uniformly for $z\in K$. Those sums are <rational functions> $\sum_jc_j/(\zeta_j-z)$ with <poles> $\zeta_j\notin K$. Thus arbitrary exterior <poles> suffice.
To move them to the prescribed set, let $A$ be the uniform closure on $K$ of <rational functions> with <poles> only in $S$. It is a closed unital algebra; it contains the coordinate function when $\infty\in S$. Define
$$
E=\{a\in\mathbb C\setminus K:(z-a)^{-1}\in A\}.
$$
This set is open relative to $\mathbb C\setminus K$. Indeed, if $a\in E$ and $|b-a|<\operatorname{dist}(a,K)$, the uniformly convergent expansion
$$
\frac1{z-b}=\sum_{j=0}^{\infty}\frac{(b-a)^j}{(z-a)^{j+1}}
$$
places the new reciprocal in $A$, using its algebra and closedness properties. It is also relatively closed: if $a_m\in E$ tends to $a\notin K$, then $(z-a_m)^{-1}$ tends uniformly on $K$ to $(z-a)^{-1}$. Each bounded component contains a finite selected <pole> and hence meets $E$. The unbounded component also meets $E$ if it contains a finite selected <pole>; if its selected <pole> is infinity, take $|a|>\sup_K|z|$ and use
$$
\frac1{z-a}=-\frac1a\sum_{j=0}^{\infty}\left(\frac za\right)^j,
$$
a uniform <polynomial> expansion. Being both open and closed and meeting every component, $E$ is all of $\mathbb C\setminus K$. Every initial Cauchy-sum reciprocal therefore belongs to $A$, and so does $f$. This proves <Runge theorem>, including its prescribed-pole assertion, rather than merely <polynomial> approximation on a special <compact set>.
For the requested <pointwise polynomial approximation of a half-plane sign>, take $n\ge2$ and
$$
K_n^+=\{x+iy:|x|\le n,\ 1/n\le y\le n\},\qquad
K_n^0=[-n,n],\qquad K_n^-=\{x+iy:|x|\le n,\ -n\le y\le-1/n\},
$$
These three <compact sets> are disjoint, and their union $K_n$ has <connected> complement: there are horizontal gaps above and below the segment, and every complementary point can be joined through these gaps or around a rectangle to the exterior of $[-n,n]^2$. Define $h_n$ to be $1,0,-1$ on disjoint open <neighborhoods> of these three sets. It is <holomorphic> on their union. By the <polynomial Runge theorem>, choose $P_n$ with $\sup_{K_n}|P_n-h_n|<1/n$.
For every fixed point with positive imaginary part, both coordinate bounds and $1/n\le\operatorname{Im}z$ eventually hold, so it lies in $K_n^+$ for all large $n$; the reflected assertion holds for negative imaginary part. Every real point eventually lies in $K_n^0$. Consequently
$$
\boxed{\lim_{n\to\infty}P_n(z)=\operatorname{sgn}(\operatorname{Im}z),\quad\operatorname{sgn}(0)=0.}
$$
The discontinuity of the limit does not contradict holomorphicity of the approximating functions: this convergence is pointwise, and is not locally uniform near the real axis.
Finally, identify $K^*$ as the <holomorphic convex hull> relative to $\Omega$. If $U$ is a component of $\widehat{\mathbb C}\setminus K$ lying entirely in $\Omega$, it cannot contain infinity and is bounded. Its boundary lies in $K$, and its closure is <compact> in $\Omega$. For any <holomorphic> $f$ on $\Omega$, the <maximum modulus principle> on $U$ gives $|f(w)|\le\sup_{\partial U}|f|\le\sup_K|f|$. Thus $U\subseteq K^*$, as is $K$ itself.
Conversely let $w\in\Omega\setminus K$ lie in a complementary component $U$ that is not entirely in $\Omega$. If it contains a finite point $a\notin\Omega$, form the <uniform algebra> on $K$ generated by constants and $(z-a)^{-1}$. The same open-and-closed reciprocal argument above, now only on the <connected component> $U$, shows that $(z-w)^{-1}$ is in its uniform closure. Its approximants have <poles> only at $a$ and are <holomorphic> on $\Omega$. If the available point outside $\Omega$ is infinity, use the <polynomial> algebra and its large-pole expansion instead. Hence in either case choose $h\in\mathcal O(\Omega)$ with
$$
\sup_{z\in K}\left|\frac1{z-w}-h(z)\right|<\frac1{2\max_{z\in K}|z-w|}.
$$
The function $F(z)=1-(z-w)h(z)$ is <holomorphic> on $\Omega$, has $F(w)=1$, and satisfies $\sup_K|F|<1/2$. It separates $w$ from the hull, so $w\notin K^*$. Combining the two inclusions proves
$$
\boxed{K^*=K\ \cup\!\!\bigcup_{\substack{U\text{ component of }\widehat{\mathbb C}\setminus K\\U\subset\Omega}}U.}
$$
Here $\mathbb P$ denotes the <Riemann sphere>; using it ensures that the unbounded component, which contains infinity, is not accidentally filled. The empty <compact set> has empty hull by the usual convention; the separation argument above concerns nonempty $K$.
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