When is a compact of the plane included in a connected compact with empty interior? The question is pretty much in the title : are there any nice conditions for a compact $K$ of the plane to be included in some other connected compact with empty interior ?
One obvious necessary condition is to have empty interior, so we can assume that it is the case. One obvious sufficient condition would be to only have finitely many connected components.
 A: That $K$ has empty interior is also sufficient. For a given nonempty compact $K$, let
$$\mathfrak{C}(K) = \left\{ C : K \subset C, \; C \text{ compact and connected} \right\}.$$
Since $\mathfrak{C}(K)$ contains a closed disk with large enough radius, it is not empty. It is partially ordered by inclusion. Next, we see that it is inductively (downward) ordered:
Let $\mathscr{C} \subset \mathfrak{C}(K)$ a chain, that is, a totally ordered subset. Let $C_0 = \bigcap \mathscr{C}$. Then $C_0$ is a compact set containing $K$, and I maintain that $C_0$ is connected: suppose there are two nonempty disjoint closed sets $F_1,F_2$ with $C_0 = F_1 \cup F_2$. Since the plane is normal, we can separate $F_1$ and $F_2$ by disjoint open neighbourhoods $U_1 \supset F_1$ and $U_2\supset F_2$. By construction,
$$\bigcap_{C\in\mathscr{C}} (C \setminus (U_1\cup U_2)) = \varnothing.$$
By the finite intersection property of compact sets, there are finitely many $C_1,\dotsc,C_k \in \mathscr{C}$ with
$$\bigcap_{\kappa = 1}^k (C_\kappa \setminus (U_1\cup U_2)) = \varnothing.$$
Since $\mathscr{C}$ is a chain, one of the $C_\kappa$, say $C_1$ is the smallest of the $C_\kappa$, and so it follows that $C_1 \setminus (U_1\cup U_2) = \varnothing$, or $C_1 \subset U_1 \cup U_2$. Since $C_1$ is connected, we have - without loss of generality - $C_1 \subset U_1$, and hence $C_0 \subset C_1 \subset U_1$, or $F_2 = C_0 \cap U_2 = \varnothing$, contradicting the assumption.
So $\mathfrak{C}(K)$ contains minimal elements by Zorn's lemma. Let $M \in \mathfrak{C}(K)$ be such a minimal element.
Suppose $M^\circ \neq \varnothing$. Then let $x\in M^\circ$, and $\varepsilon > 0$ such that $D_\varepsilon(x) \subset M$. Since $K$ has empty interior, there is an $y \in D_\varepsilon(x)\setminus K$. Choose $\delta > 0$ such that $D_\delta(y)  \cap K = \varnothing$ and $D_{2\delta}(y) \subset D_\varepsilon(x)$.
Then $M \setminus D_\delta(y)$ is a compact set containing $K$, it is a proper subset of $M$, and it is connected, which contradicts the minimality of $M$.
