I'm essentially reopening this old question of Ricky Demer which was never fully answered.
Essentially the original question: Suppose we have a topological field $F$ which is complete, Hausdorff, and non-discrete, and we put a Hausdorff topology on $F^n$ so as to make it a topological vector space over $F$; is is this topology necessarily the product topology (and hence complete, and hence closed in anything it embeds in)?
As the link shows, the answer is yes if $\tau$ comes from an absolute value on $F$, and it's easy to see the same argument works if it comes from a field ordering on $F$.
Note that the argument there shows that this question reduces to the following lemma:
Suppose we have a topological field $(F,\tau)$ which is Hausdorff and non-discrete, and we give $F$ a second topology $\tau'$, which is also Hausdorff, such that $(F,\tau')$ is a topological vector space over $(F,\tau)$. Does this force $\tau=\tau'$? What if we assume that $(F,\tau)$ is complete?
(I'm isolating completeness as a separate, possibly-unnecessary condition because the argument there only uses completeness in the reduction to the lemma, not in proving the lemma for valued fields.)
Related to this question in that one way to come up with a counterexample for both simultaneously would be to find a field with two (nondiscrete, Hausdorff) topologies with one strictly finer than the other.