Note: Please do not give a solution; I am curious to understand why my solution is incorrect, and would prefer guidance to help me complete the question myself. Thank you.
Problem
Show that if $f$ is an infinitely times differentiable function on $\mathbb{R}$ such that for every $x\in\mathbb{R}$ there exists $n\geq 1$ with $f^{(n)}(x)=0$ then there exists $N\geq 1$ such that for every $x\in\mathbb{R}$ we have $f^{(N)}(x)=0$.
My Proof
My proof relies on Baire's Theorem. However I am unable to show that $C^{\infty}(\mathbb{R})$ is a metric space (I would appreciate some guidance in the form of a hint on this). Hence, for the time being I have assumed this to be true and continued with the proof as follows.
For every $x\in\mathbb{R}$ define the set $O_{n}:=\{f^{(n)}(x)\in C^{\infty}(\mathbb{R})|f^{(n)}(x)\neq 0\text{ and }f^{(n)}(x)\rightarrow_{n\rightarrow\infty}0\}$. Then every $O_{n}$ is open and dense in $C^{\infty}(\mathbb{R})$. Hence, by Baire's Theorem, $\bigcap O_{n}$ is dense in $C^{\infty}(\mathbb{R})$. Now we define $N:=\inf\{n\geq 1|f^{(n)}(x)=0, \forall f^{(n)}(x)\in\bigcap O_{n}\}$ so that $f^{(N)}(x)=0$ for every $x\in\mathbb{R}$.
Second Attempt
Assume $f\in\mathbb{C}^{\infty}$ then for every $x\in\mathbb{R}$ there exists $n\geq 1$ such that $f^{(n)}(x)=0$. Then for every $n\geq 1$ define, \begin{align} X_{n}:=\{x\in\mathbb{R}|f^{(n)}(x)=0\}. \end{align} Since $\mathbb{R}$ is a complete metric space, Baire's Theorem tells us there exists $N\in\mathbb{N}$ such that $X_{N}=\{x\in\mathbb{R}|f^{(N)}(x)=0\}$ is non-empty.
This is great progress, however I am not sure where to continue from here. How do I extend $X_{N}$ so that $X_{N}=\mathbb{R}$?