Suppose we have a compact \(K \subseteq \mathbb{C} \). We wish to show that the boundary of each path component of \(\mathbb{C} \setminus K \) is a subset of \(K\). Consider a path component \(P\), where \(\partial P \) is the collection of boundary points of \(P\). By definition, \( \forall q \in \partial P, q \notin P\). Thus \(q \in K \lor q\in L\), where \(L\) is a different path component of \(\mathbb{C} \setminus K\). Suppose \(q \in L\). We know that \( \mathbb{C} \setminus K\) is open, as its complement (\(K\)) is closed, as it is a compact subset of a metric space. Thus \(\exists \epsilon > 0 \ s.t. \ B\left(q, \epsilon\right) \subseteq \mathbb{C} \setminus K\). As \(q\) is a boundary point of \(P\), \( P \cap B\left(q, \epsilon\right) \neq \emptyset\). As \(B\left(q, \epsilon\right) \subseteq \mathbb{C} \setminus K\), there is a path between \(q\) and another point \(\hat q\), where \(\hat q \in B\left(q, \epsilon\right) \cap P\). This follows because \(B\left(q, \epsilon\right) \cap K \ = \emptyset\). As \(q\) and \(\hat q\) are path connected, they must be in the same path component by definition. This is a contradiction, so \(q \in K\). Thus, \(\forall q \in \partial P, q \in K\), so \(\partial P \subseteq K\).