If $M$ is a smooth manifold and $S\subset M$ is an embedded submanifold that is not closed in $M$, then can we construct a smooth function $f\in C^{\infty}(M)$ that is non-zero everywhere on $S$ and equal to $0$ atleast at one point outside $S$?.
This question came up in my attemp of solving problem 5.18 (b) from Lee's book "Introduction to smooth manifold" second edition. The problem from the book is:
Suppose $M$ is a smooth manifold and $S\subset M$ is a smooth submanifold.
(a) Show that $S$ is embedded if and only if every $f\in C^{\infty}(S)$ has a smooth extension to a neighborhood of $S$ in $M$. [Hint: if $S$ is not embedded, let $p \in S$ be a point that is not in the domain of any slice chart. Let $U$ be a neighborhood of $p$ in $S$ that is embedded, and consider a function $f\in C^{\infty}(S)$ that is supported in $U$ and equal to $1$ at $p$.]
(b) Show that $S$ is properly embedded if and only if every $f\in C^{\infty}(S)$ has a smooth extension to all of $M$.
For part (b) I am struggling to prove that $S$ is properly embedded if every $f\in C^{\infty}(S)$ has a smooth extension to all of $M$.
My attemp: Suppose $S$ is not properly embedded then it is not closed in $M$, so there exists $p\in \overline{S}$ such that $p\notin S $. Now if I am able to construct $f\in C^{\infty}(M)$ such that $f(p)=0$ and $f\neq 0$ on $S$, we have that $g=\frac{1}{f} \in C^{\infty}(S)$ that cannot be extended to a smooth function on whole of $M$.