My problem is with the part where he uses the pasting lemma to conclude that $h$ is continuous.
The pasting lemma requires that both sets on which the corresponding functions are defined be either open or closed in the whole space $X$ which is $R^2$ in this case.
$A$ is a compact subspace of $R^2 – 0$ and hence is closed in it, but that doesn't guarantee it is closed in $R^2$.
$C$ and $D$ are both open in $R^2 – A$ because they are components and the space is locally path connected, but I haven't been able to prove anything about whether they are open or closed $R^2$.
It looks like a trivial detail among the main points in the proof so I must be missing something here.
Best Answer
Compactness of a set is a topological property: it is independent of the space in which that set is embedded. $A$ is a compact subset of $\Bbb R^2\setminus\{\bf0\}$, so it is compact as a space and therefore compact as a subset of any space in which it is embedded. In particular, it is a compact subset of $\Bbb R^2$.
$C$ and $D$ are open subsets of $\Bbb R^2\setminus A$, which is open in $\Bbb R^2$, so $C$ and $D$ are open in $\Bbb R^2$. Specifically, for $C$ there is an open $U$ in $\Bbb R^2$ such that $C=U\cap(\Bbb R^2\setminus A)$, but that means that $C$ is the intersection of two open sets in $\Bbb R^2$ and hence is itself open in $\Bbb R^2$. The argument for $D$ is the same.