Proof:

Let $h$ be the homeomorphism between $U$ and $V$ and $p \in U$. We have to show that $h(p)$ is an interior point of $V$. Let $K$ be a closed ball centered at $p$ and contained in $U$ so that $K^{\prime} = h(K)$ is a compact subset of $V$ containing $q = h(p)$. Let $B$ be the (topological) boundary of $K$ and $B^{\prime} = h(B)$. We regard $U$ and $V$ as subsets of $S^n$. By theorem (41.2), $S^n - K^{\prime}$ is path connected and $S^n - B^{\prime}$ has two path components. However since the union

$$S^n - B^{\prime} = (S^n - K^{\prime})\cup (K^{\prime} - B^{\prime})$$

is a disjoint union of connected sets, the pieces $S^n - K^{\prime}$ and are the components of $S^n - B^{\prime}$. Hence they are both open in $S^n - B^{\prime}$ (why?) and hence are open in $S^n$. The piece is then an open subset of $S^n$ containing $q$ and since we see that $q$ is an interior point of $V$. $\square$