= Solution
For $x$ in a <compact Hausdorff space> $K$, its <quasi-component> $Q$ is the intersection of all <clopen sets> containing $x$. It is closed and therefore <compact>. Suppose $Q=A\sqcup B$ is a <separation of a topological space> into two nonempty relatively <closed sets>, with $x\in A$. The sets $A,B$ are <compact> and hence closed in $K$. <Normality of a compact Hausdorff space> gives disjoint <open neighborhoods> $U,V$ of $A,B$.
The <compact set> $K\setminus(U\cup V)$ is disjoint from $Q$. For each <point> $y$ of that set choose a <clopen> $C_y$ containing $x$ but not $y$. Finitely many complements cover $K\setminus(U\cup V)$, so their corresponding intersection $C$ is <clopen>, contains $Q$, and is contained in $U\cup V$. Now $C\cap U$ is <clopen> in $C$, hence <clopen> in $K$, and contains $x$ while missing $B$. This contradicts the definition of $Q$. Thus \b[$Q$ is <connected>]. Any <connected set> containing $x$ lies in every <clopen> <neighborhood> of $x$, so the <connected component> of $x$ equals $Q$.
For the second assertion, $D$ is closed in the ambient <Hausdorff space> $X$, and is itself <compact> <Hausdorff>. Let $C$ be a <connected component> of $D$ disjoint from $\partial_XD$. Choose $p\in C$. By the first assertion $C$ is the <quasi-component> of $p$ in $D$. For each $y\in\partial_XD$ choose a <clopen> subset of $D$ containing $p$ but not $y$. <Compactness> of $\partial_XD\subseteq D$ gives a <finite intersection> $H$ of these sets which contains $C$ and avoids the whole <boundary>. Thus $H\subseteq\operatorname{int}_XD$. It is closed in $X$, because it is closed in the <closed set> $D$, and is open in $X$, because it is relatively open in $D$ and contained in its <interior>. It is nonempty and proper, contradicting <connectedness> of $X$. Hence \b[every <connected component> of $D$ meets its <boundary>].
Finally suppose a closed subset $A$ of the square has the <point>-separating <clopen> property specified in the paper. Every nonempty relatively open subset of the square contains a nontrivial <line segment>: it contains an <interior> <point> of the square and a small ball about that <point>. If such a segment lay in $A$, a <clopen> subset of $A$ separating its endpoints would induce a <separation of a topological space> of the <connected> segment. This is impossible. Therefore $A$ contains no nonempty relatively open subset of the square. Being closed, it is \b[<nowhere dense> in the square], and also <nowhere dense> in $\mathbb R^2$.
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