I think I'm misunderstanding what is being asked of me in the following question:
The space $C^{\infty}(\mathbb{R})$ has a Fréchet space topology with respect to which $f_n \rightarrow f$ iff $f^{(k)}_n \rightarrow f^{(k)}$ uniformly on all compact subsets ($k$, non-negative, is the order of derivative).
To show the backwards implication I need to show that convergence ($f^{(k)}_n \rightarrow f^{(k)}$ uniformly on all compact subsets) give me seminorms, there are countably many such semi norms, and these seminorms give me a complete Hausdorff TVS. My book gives me some facts to help me in this direction.
I'm confused by the forward implication though - if endow the space of $C^{\infty}(\mathbb{R})$ with a topology given by countable seminorms which makes $C^{\infty}(\mathbb{R})$ a complete Hausdorff TVS -- how do I know this implies $f^{(k)}_n \rightarrow f^{(k)}$ uniformly on all compact subsets? What if the seminorms are really strange and have nothing to do with the derivatives of $f$ over compact subsets?
The question doesn't exclude that there are different topologies on $C^\infty(\mathbb{R})$ such that $C^\infty(\mathbb{R})$ is a Frechet space with this topology. However, since the open mapping theorem is also true for Frechet spaces, we cannot have a finer (or weaker) topology (as the usual one) on $C^\infty(\mathbb{R})$ in which $C^\infty(\mathbb{R})$ is also a Frechet space and $f^{(k)}_n \rightarrow f^{(k)}$ uniformly on compact sets for each $k \in \mathbb{N}$.
One can proof that $C^\infty(\mathbb{R})$ has dimension $2^{\aleph_0}$ in the sense of Hamel basis. Thus you can take e.g. the Banach-space $L^1(\mathbb{R})$ and transport the norm to $C^\infty(\mathbb{R})$. With these norm $C^\infty(\mathbb{R})$ is a Banach space. Since the topology of local convergence of all derivates is not normable, this weird topology is different than the usual one.