Every finite dimensional vector space over $\mathbb{R}$ or $\mathbb{C}$ has a unique topology that makes addition and scalar multiplication continuous. Is the same true of finite dimensional vector spaces over any locally compact field?
I assume that $F$ has a locally compact topology with respect to which addition, multiplication, and additive inverse are all continuous. If we also need to assume that multiplicative inverse is continuous on $F\setminus\{0\}$, that is fine. We can also assume the topology is metrizable if that helps.
The multiplicative inverse map is automatically continuous by $F$ being locally compact Hausdorff. (We cannot drop locally compactness, this is the difference between Banach field and nonarchimedean field in Kedlaya's paper.)
If $F$ is discrete, then definitely no. Take $F=\mathbb R$ with discrete topology, and $V=\mathbb R^n$ with the usual Euclidean topology. Otherwise (if $F$ is non-discrete), $F$ must be a local field, hence a complete non-discrete valued division ring. This can be found in e.g. Weil's Basic Number Theory, roughly the automorphism $x\mapsto ax$ of $F$ scales the Haar measure which can be used to define the valuation of $a\in F^{\times}$.
If the topology on $V$ is allowed to be non-Hausdorff, it can be the trivial topology. Otherwise, the answer is yes. This has been done at different levels of generality.
For local fields: Chapter 1, Section 2, Corollary 1 of Theorem 3 of Basic Number Theory.
For complete non-discrete valued division ring: Chapter 1, Section 2.3 "Vector subspaces of finite dimension" of Bourbaki's Topological Vector Spaces. Here is a sketch.
For complete non-discrete division algebra: MathOverFlow discussion. This can avoid establishing local fields can be valued.