Every manifold is paracompact. I tried:

$M$ is an $n$--manifold with open covering $U_\alpha$ and $\varphi_\alpha$ local homeomorphisms; $\varphi_\alpha (U_\alpha)$ are open in $\mathbb R^n$. Adding $B(x, \varepsilon)$ for $x \in (\bigcup_\alpha \varphi_\alpha (U_\alpha))^c$ yields an open covering of $\mathbb R^n$. $\mathbb R^n$ is paracompact hence there is a refinement $V_\alpha$. We discard $V_\alpha \subseteq B(x,\varepsilon)$ and observe that $\varphi_\alpha^{-1}(V_\alpha)$ are a refinement of $U_\alpha$. Fix $p \in M$ and $\alpha_0$ with $p \in U_{\alpha_0}$. Then there is an open nbhd $N$ of $\varphi_{\alpha_0} (p)$ such that $N$ intersects only finitely many $V_\alpha$. Let $N' = \varphi_{\alpha_0}^{-1}(N \cap \varphi_{\alpha_0} (U_{\alpha_0}))$. Then $N'$ is an open nbhd of $p$.

My intended finish was "$N'$ only intersects finitely many $\varphi_\alpha^{-1}(V_\alpha)$". Alas, it appears that one cannot argue like this since $\varphi_\alpha$ and $\varphi_{\alpha_0}$ map $\varphi_\alpha^{-1}(V_\alpha)$ to different sets. How to salvage the proof? Thank you.