In James W. Vick's "Homology Theory: An Introduction to Algebraic Topology," Theorem 1 of Appendix II states that if $M$ is a compact $n$-manifold without boundary, then there is some $m \in \mathbb{N}$ such that $M$ embeds homemorphically into $\mathbb{R}^m$. His proof proceeds as follows.
Let $N \in S^n$ denote the North Pole. Cover $M$ with finitely many open sets $U_1, ..., U_l$, each equipped with a homeomorphism $h_i:U_i \rightarrow S^n - \{N\}$. Define $\overline{h}_i: M \rightarrow S^n$ by $\overline{h}_i(x)=h_i(x)$ for $x\in U_i$, and $\overline{h}_i(x)=N$ for $x\in M-U_i$. Vick claims the maps $\overline{h}_i$ are continuous, so we get a homeomorphic embedding $M \rightarrow (\mathbb{R}^{n+1})^l$ defined by $x \mapsto (\overline{h}_1(x), ..., \overline{h}_l(x))$.
Here is my question: how do we know $\overline{h}$ is continuous? I know this fact must rely on Hausdorffness because of the following counterexample. Take $M$ to be the circle with two South Poles, and obtain $U$ from $M$ by removing the North Pole and one of the South Poles. There is an obvious choice of homeomorphism $h:U \rightarrow S^1 - \{N\}$, and clearly the extension $\overline{h}$ is not continuous.
So how does the continuity of $\overline{h}$ follow from the Hausdorffness of $M$?
$\overline h_i$ is continuous at all points $x\in U_i$ because its restriction to the neighbourhood $U_i$ of $x$, which is $h_i$, is continuous at $x$.
So let $x\notin U_i$. We need to prove that $\overline h_i$ is continuous at $x$, so let $V\subseteq S^n$ be an open neighbourhood of $\overline h_i(x)=N$. We know that $S^n\setminus V$ is a compact subset of $S^n\setminus\{N\}$ and therefore that $\overline h_i^{-1}(S^n\setminus V)=h_i^{-1}(S^n\setminus V)$ is a compact subset of $U_i$ (because $h_i$ is a homeomorphism), and therefore of $M$. By the set-theoretic identity $f^{-1}(A)=(f^{-1}(A^\complement))^\complement$, we have $\overline h_i^{-1}(V)=M\setminus\overline h_i^{-1}(S^n\setminus V)$. So $\overline h_i^{-1}(V)$ is the complement of a compact subset, and therefore it's open.