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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 $K \subseteq B \subseteq A \subseteq V$ ?

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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.

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    Why does $K'$ has finitely many connected components ? and why is $B$ open and connected ? – 2017-02-14
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    Oops, my definition of $B$ was mistaken, I fixed it, it's a union of open balls (not closed balls). And I added a few reasons. – 2017-02-14
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    I just have only difficulty in understanding why $B$ is connected ... why does being a union of connected sets with each set containing a point from a common connected set ( the points may not be same) imply that the union is connected ? – 2017-02-14
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    For two sets this is simple (and may be found in a proposition in Munkres textbook on topology). For a finite number of sets, apply induction. – 2017-02-14
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    the result I know is that if two connected sets have non-empty intersection then their union is connected . But here why should any two balls should intersect ? – 2017-02-15
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    There is no assertion that any two of the balls $B(y_j,s_j/2)$ intersect. What *is* true is that any of the balls $B(y_j,s_j/2)$ intersects $K''$, namely $y_j \in B(y_j,s_j/2) \cap K''$. Can you figure out how the induction proof works now? – 2017-02-15