In Lee's Introduction to Smooth Manifolds, problem 5-18a has my brain making uncomfortable cracking noises. The question is to show that $S$ is an embedded submanifold of $M$ if and only if every $f\in C^{\infty}(S)$ has a smooth extension to an open neighborhood $U$ of $S$. The first implication is quoting a theorem, but the other one "$\Leftarrow$" I've been unsuccessful at.
The book gives the following 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 $p$.
I'm having trouble connecting the dots with this hint. It feels like I'm missing something very obvious. I've also tried another method, but that one's very sketchy and probably not legit:
Since every $f\in C^{\infty}(S)$ has an extension to a neighborhood $U$ in $M$, we can also consider these $f$ as the restriction of some $F\in C^{\infty}(U)$ to $S$. Pick $n$ linearly independent such $F$, denoted $F_1, \ldots, F_n$. Let $\dim(S)=k$. We can assume $k<n$ since otherwise $S$ would be embedded. Hence $S$ is closed in $M$ w.r.t. the subspace topology, and there exists a function $\phi\in C^{\infty}(U)$ such that $\phi|_S=1$ and $\text{supp }\phi \subseteq U$. Then the function:
$$\psi = (F_1, \ldots, F_k, (1-\phi)F_{k+1}, \ldots, (1-\phi)F_n) $$
is a slice chart of $S$, which can only exist if $S$ is an embedding.
I'd like to know if my proof was legit (and if not, if it's a legit strategy to begin with), but more than that I want to know what Lee was alluding to with his hint. I've been staring at this problem for days and it's driving me crazy.
Just follow the hint as follows:
Let $\tilde{f}$ be a smooth extension of $f$ on a neighborhood $\tilde{U}$ of $S$. Consider the following two sets:
$$ V := f^{-1}((0, \infty)) \\ W := \tilde{f}^{-1}((0, \infty)) $$ Since $f \in C^\infty(S)$ and $\tilde{f} \in C^\infty(\tilde{U})$, $V$ is open in $S$ and $W$ is open in $\tilde{U}$, thus open in $M$. By construction of $f$ and $\tilde{f}$, both $V$ and $W$ contain $p$ and $W\cap S = V$. So there exists a k-slice chart around $p$. It contradicts with choice of $p$.