Is it true that any separated quasi-compact rigid-analytic variety embeds into a proper one?
For my purpose, the base field is a $p$-adic number field.
I have seen Huber's universal compactification in Etale Cohomology of Rigid Analytic Varieties and Adic spaces, but I didn't quite understand its statement. Huber states the existence of a universal compactification, which seems not exist in rigid varieties. How to understand this?
Notation: $K=\mathbf{Q}_p$, $K^+$ is its valuation ring and $k$ is its residue field.
A rigid space is an adic space locally of finite type over ${\rm Spa}(K, K^+)$.
(1) I think it probably depends whether by "embeds" you mean an open immersion or a locally closed one. (A locally closed subspace of a rigid space might not be open in its Zariski closure.) But I don't know the answer in either case.
If $X$ is smooth and affinoid, I think it follows from Elkik's theorem that $X$ is an analytic open in the analytification of a smooth affine algebraic variety. Since that variety can be compactified, the answer is positive in this case.
(2) In the context of your question, it is useful to know that if $X$ embeds as an open in a (partially) proper $Y$, then Huber's universal compactification of $X$ is homeomorphic to the closure $\overline{X}$ of $X$ in $Y$.
Since on the level of Berkovich spaces, $X^{\rm Berk}$ and $Y^{\rm Berk}$ are compact, the former is closed in the latter, and it follows that $\overline{X}$ and $X$ have the same maximal (rank one) points.
For example, if $X$ is the affinoid unit disc and $Y=\mathbf{P}^1$, then $\overline{X} = X \cup \{\zeta\}$ where $\zeta$ is the point corresponding to the rank two valuation on $K[x]$ which is the composite of the Gauss norm (with residue field $k(x)$) and the valuation "order of vanishing at infinity" of $k(x)$. Note that this valuation is not $\leq 1$ on $K^+[x]$, so does not define a point of ${\rm Spa}(K[x], K^+[x])$.
In general, an affinoid rigid space is of the form $X={\rm Spa}(A, A^\circ)$ where $A^\circ$ is the subring of powerbounded elements. Its universal compactification is then ${\rm Spa}(A, A^+_{\rm min})$ where $A^+_{\rm min}$ is the smallest ring of integral elements containing $K^+$, namely the integral closure of the subring of $A$ generated by $K^+$ and the topologically nilpotent elements $A^{\circ\circ}$.
