Let $X$ be a vector space and equip it with the weak topology induced by all linear functionals $X\rightarrow \mathbb{C}$. I am wondering if this topology is so strong that it is determined on finite dimensional subspaces. More precisely, do we have that $C\subseteq X$ is closed if and only if $C\cap Y$ is closed for any finite dimensional subspace $Y$?
I did some computations that seem to suggest this. For example, suppose $\{x_n\}$ is a sequence which converges to $x$ in this topology. Then $\{x_n: x\in \mathbb{N}\}$ must be contained in some finite dimensional subspace. Indeed, if not, then we can find a subsequence $\{x_{n_k}\}$ such that $x_{n_k}\notin \text{Span}\{x_m: m<n_k\}$ and consider a linear functional $\phi$ that sends $x_{n_k}$ to $k$, for each $k$. Then $\phi(x_{n_k})\rightarrow \infty$ and $x_n$ cannot converge to $x$. I have tried doing similar tricks for nets, but am really struggling to make it work.
For the record, since any seminorm must be continuous on $X$, we know that the topology is stronger than any norm topology on $X$. I have tried to consult Kelley's and Namioka's 'Linear Topological Spaces', but could not find anything about this exact topology. However this is such an 'obvious' topology to look at that I figure it must be known.
Contrary to what you say, this topology is not stronger than norm topologies. Indeed, any basic open neighborhood of $0$ in this (or any) weak topology must contain a linear subspace of $X$ of finite codimension (namely, the intersection of the kernels of the finitely many functionals that define the neighborhood). So if $X$ is infinite-dimensional, this weak topology does not contain any norm topology, since the unit ball in any norm topology cannot be open in the weak topology. It follows also that the weak topology is not determined by finite-dimensional subspaces.