The weak approximation theorem states that given a field $F$ and nontrivial inequivalent absolute values $|\cdot|_1,\ldots,|\cdot|_n,$ and letting $F_i$ denote $F$ with the topology from $|\cdot|_i$, then the diagonal in $F_1 \times \ldots \times F_n$ is dense.

So suppose now we have the same setup, except now the topologies don't necessarily come from absolute values (but do make $F$ a topological field, and are all still distinct and nondiscrete and of course Hausdorff). Does the result still hold?

Related to this older 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.