Let $k$ be a field complete with respect to a non-archimedean absolute value, and $X$ a separated algebraic space of finite type over $k$.

If $X$ is a scheme then $X(k)$ inherits a natural (Hausdorff) topology from $k$ (uniquely determined by functoriality, compatibility with open immersion, closed immersions, fiber products, and the special case of the affine line), locally compact when $k$ is locally compact. There is even a natural $k$-analytic manifold structure if $X$ is smooth. This can be seen in (at least) two ways: algebraically by using Zariski-open covers by affines, and for smooth case take them to be "standard etale" over affine spaces (and then the $k$-analytic inverse function theorem can be applied), or analytically by identifying $X(k)$ with $X^{\rm{an}}(k)$ for the analytification $X^{\rm{an}}$ in the sense of rigid-analytic spaces (so then affinoid open covering of $X^{\rm{an}}$ does the job).

Now suppose $X$ is not a scheme (as often happens with moduli spaces which only exist as algebraic spaces; e.g., Rapoport's thesis, etc.). The 2nd method above can be used, but constructing $X^{\rm{an}}$ as a rigid-analytic space is really hard (the only method I know is to take a very long detour through Berkovich spaces to make the required rigid-analytic quotients from an etale scheme chart for $X$).

My question is this: is there a known simple procedure (much as the method in the scheme case is simple), bypassing the use of $X^{\rm{an}}$, to make a functorial Hausdorff topology on $X(k)$ (and $k$-analytic manifold structure when $X$ is smooth) which is locally compact when $k$ is and recovers the usual topology when $X$ is a scheme and shares the same basic properties (good behavior for open immersions, closed immersions, and fiber products; carries etale maps to local homeomorphisms as well as local $k$-analytic isomorphisms in case $X$ is smooth; and $X(k) \rightarrow X(k')$ is a closed embedding any extension $k'/k$ of such fields)?

A natural idea for the topology aspect is to choose an etale scheme cover $U \rightarrow X$ and identify $X(k_s)$ with a quotient of $U(k_s)$ as a set. Give it the quotient topology of the natural topology on $U(k_s)$ arising from the topology on $k_s$, and give $X(k)$ the subspace topology from $X(k_s)$. This does give the right answer (same as via the rigid-analytic method), but the only way I see this is locally compact when $k$ is (noting that $X(k_s)$ is essentially never locally compact) and functorial is to invoke comparison with the rigid-analytic method. (Otherwise I get bogged down in rising chains of finite Galois extensions and it feels like it becomes a mess. Doing the $k$-analytic manifold structure for smooth $X$ in this way also seems to become quite unpleasant. Note, by the way, that $k_s$ is not complete, so can't speak of $k_s$-analytic manifolds.) So this idea doesn't seem to provide an answer. Or maybe I am missing a simple trick?

I am happy to restrict to the case of locally compact $k$ (though allowing completed algebraic closures is perfectly interesting), but do not want to restrict to characteristic 0 (though ideas in that case are appreciated too).