= Solution
For $\lambda\notin K$, write $g_\lambda(z)=(z-\lambda)^{-1}$. If $\lambda\in\Lambda$ and
$$
|\mu-\lambda|<\operatorname{dist}(\lambda,K),
$$
then
$$
g_\mu
=\frac{g_\lambda}{1-(\mu-\lambda)g_\lambda}
=\sum_{n=0}^{\infty}(\mu-\lambda)^ng_\lambda^{n+1}
$$
uniformly on $K$. Uniformly polynomially approximable functions form an algebra closed under uniform limits, so $\mu\in\Lambda$. Hence $\Lambda$ is open.
The set $\Lambda$ is also closed relative to $\mathbb C\setminus K$. Indeed, if $\lambda_j\in\Lambda$ and $\lambda_j\to\lambda\notin K$, then
$$
\|g_{\lambda_j}-g_\lambda\|_\infty
\leq
\frac{|\lambda_j-\lambda|}
{\operatorname{dist}(\lambda_j,K)\operatorname{dist}(\lambda,K)}
\longrightarrow0,
$$
so $g_\lambda$ is again uniformly approximable by polynomials. Since $K$ is closed, every point of $\Gamma$ has a neighbourhood in $\mathbb C\setminus K$; relative closedness of $\Lambda$ then gives a smaller neighbourhood disjoint from $\Lambda$. Thus \b[both $\Lambda$ and $\Gamma$ are open].
It is not always true that $\Lambda$ is nonempty. Take $K=\mathbb R$. Any polynomial uniformly close on $\mathbb R$ to a bounded resolvent $g_\lambda$, with $\lambda\notin\mathbb R$, must itself be bounded on $\mathbb R$, and hence must be constant. Uniform limits of constants are constant, whereas $g_\lambda$ is not. Therefore
$$
\boxed{K=\mathbb R\quad\Longrightarrow\quad\Lambda=\varnothing.}
$$
Boundedness of $K$ does not force $\Gamma$ to be empty either. Take $K=\{z:|z|=1\}$ and $\lambda=0$. If polynomials $p_n$ converged uniformly to $1/z$ on $K$, then <uniform convergence and contour integration> would give
$$
0=\lim_{n\to\infty}\oint_Kp_n(z)\,dz
=\oint_K\frac{dz}{z}=2\pi i,
$$
a contradiction. Hence
$$
\boxed{K=\{z:|z|=1\}\quad\Longrightarrow\quad0\in\Gamma.}
$$
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