The following result due to Sunyer y Balaguer and Corominas is somewhat of a classic:
Let $f: \mathbb{R} \to \mathbb{R}$ be of class $C^{\infty}$. For every $x \in \mathbb{R}$, there exists $n_x \in \mathbb{N^*}$ such that $f^{(n_x)}(x)=0$. Then $f$ is a polynomial.
Here $\mathbb{N^*}$ denotes the set of non-negative integers.
This was an exercise on my test, where you are given the steps of the proof and you must fill in each step. Proving each step individually is not too hard, but there were too many steps and so I lost sight of the big picture.
-The first step is usually to consider the set $S$ of real $x$ such that $f$ coincides with a polynomial in some neighborhood of $x$. (What's the intuition for considering this set?)
-Then you prove that $S$ is open and the complement of $S$ has no isolated points, which is easy enough.
-Then $S$ is an at-most-coutable union of disjoint open intervals, and we can show that on each of these intervals, $f$ coincides with a polynomial.
-After that, we suppose that $S$ is not equal to $\mathbb{R}$ (or we'd be done), and we deduce a contradiction. It is this part of the proof, which uses Baire's category theorem where I got lost in the details.
Can please someone provide the grand lines of the proof and the intuition?