A property regarding open domain in complex plane Let $V$ be an open connected subset of $\mathbb R^2$ , then is it true that for every compact set $K \subseteq V$ , there exist a compact set $A$ and an open connected set $B \subseteq \mathbb R^2$ such that $K \subseteq B \subseteq A \subseteq V$ ? 
 A: Yes, this is true. 
For the proof, the first step is to choose a compact $K'$ such that $K \subset K' \subset V$ and $K'$ has only finitely many components. To do this, use compactness to obtain finitely many open balls $B(x_i,r_i) \subset V$ centered on points $x_i \in K$ such that the half-radius balls $B(x_i,r_i/2)$ cover $K$, and then take 
$$K' = \bigcup_i \overline{B(x_i,r_i/2)}
$$
Since $K'$ is a union of finitely many connected sets, $K'$ has finitely many components.
Next, choose a compact $K''$ such that $K' \subset K'' \subset V$ and $K''$ is connected: enumerate the components $K'=K'_0 \cup K'_1 \cup\cdots\cup K'_M$, for each $m=1,\ldots,M$ let $\gamma$ be the image of a path in $V$ connecting a point of $K'_0$ to a point of $K'_m$, and let 
$$K'' = K' \cup \bigcup_{m=1}^M \text{image}(\gamma_m)
$$
Finally, take finitely many open balls $B(y_j,s_j) \subset V$ centered on points $y_j \in K''$ such that the half-radius balls $B(y_j,s_j/2)$ cover $K''$, and take
$$B = \bigcup_j B(y_j,s_j/2) \quad\text{and}\quad A = \overline B
$$
Since $B$ is a union of open sets, it is open. Also, since $B$ is a union of a connected set (namely $K''$) with a collection of connected sets each containing a point of $K''$, $B$ is connected.
