Every complete topological vector space is metrizable

Question

I am looking for a proof of the following result:

If $(X,\tau)$ is a complete topological vector space, then $\tau$ is generated by a complete distance wich is invariant by translations.

My attempt is based on a similar result that I has found in https://en.wikipedia.org/wiki/Fr%C3%A9chet_space#Constructing_Fr%C3%A9chet_spaces. However, why can we ensure the existence of such $\|\cdot\|_k$? and, is it true that $\tau$ is generated by such distance (I don't see any relation between both topologies)?

Answer 1

Your claimed result is false: $\mathbb{R}^{\omega_1}$ in the product topology is a locally convex complete TVS (in the sense mentioned in the comments) but is not metrisable at all.