The following lemma is my question. (cf GTM218, Introduction to Smooth manifold)
I can prove (b) using partion of unity as follows:
$Proof$ for any $p \in S$ choose a slice chart $W_p$ centered at $p$ and extend the restriction $f|W_p \cap S$ to some $f_p \in C^{\infty} (W_p)$. Now ,take a partition of unity $\{\phi_P: p \in S\}\cup\{\chi\}$, which issubordinate to the covering $\{W_P: p \in S\}\cup\{M \smallsetminus S\}$ of $M$( in this book, the condition $S$ is properly embedded is equivalent to the condition that $S$ is closed and embedded). Then ,we set $$ \tilde f = \sum_{p \in S} \phi_p f_p$$
One can check this function is a extension for $f$.
However, this method cannot be applied to show (a), where $S$ is not assumed to be closed.
Best Answer
If I haven't missed something, the only thing you cannot do in the case of (a) is adding $M\setminus S$ to the covering. Despite this, $\{W_p\mid p\in S\}$ is still a covering of some neighborhood $$U=\bigcup_{p\in S}W_p$$ of $S$. Therefore, simply taking a partition of unity $(\phi_p)_{p\in S}$ subordinate to this cover allows you to construct the desired function $f\in C^{\infty}(U)$ in exactly the same way as you did for point (b).
Added: In fact, $S$ is properly embedded in $U$. It is embedded, since each point lies in a slice chart, so we only have to see that it is closed. To see this, notice that $W_p\cap S$ is closed in $W_p$ (since $W_p$ is a slice chart). So $W_p\setminus S$ is open in $W_p$, which is open in $M$. Therefore $W_p\setminus S$ is open in $M$ for each $p$, so $$\bigcup_{p\in S}(W_p\setminus S) = U\setminus S$$ is open in $M$. But $U\setminus S\subseteq U$, so $U\setminus S$ is in fact open in $U$. Therefore, $S$ is closed in $U$.