A function $f$ which is 1 on $V$ and 0 on $U^c$

Question

I want to prove this proposition.

Let $M$ be a manifold. For any $p\in M$ and open neighborhood $p \in U \subset M$, there exists a open neighborhood $V$ of $p$ and $C^{\infty}$ function $f$ on $M$ such that $p\in V \subset \bar{V}\subset U$, $0\leq f\leq 1, f = 1 $ on $V$ and $f=0$ on $U^c.$

My textbook proves as follows:Let $U'\subset U$ be a open neighborhood of $p$ which is homeomorphic to an open set in $\mathbb{R^n}$. There exists open sets $W, V$ and $C^\infty$ function $f$ on $U'$ such that $p\in W\subset \bar{W}\subset V \subset U'$, $f=1$ on $\bar{W}$, $f=0$ on $V^c$, $0\lt f \lt 1$ on $V-\bar{W}$. Defining $f=0$ on $M-V$, we obtain a required function $\bar{f}$.

I think that the fact that $\bar{f}$ is $C^\infty$ needs a proof using Hausdorffness of $M$, because if $M$ is not Hausdorff, it may be false. For example, let $p$ be a origin of the line with two origin and $q$ be the other origin, $\bar{f}$ is not continuous at $q$ since $\bar{f}=1$ near q and $\bar{f}(q)=0$.

Answer 1

You can actually construct these explicitly. Try modifying the example given here, on page 15.