In case you're still interested: Bhatt recently proved that for any Tate-Huber pair $(A,A^+)$, the topological space $\mathrm{Spa}(A,A^+)$ is homeomorphic to an inverse limit of admissible blowups in the expected manner (Theorem 8.1.2 here). For fun I'll summarize the construction (which probably won't seem too surprising). Fix a topologically nilpotent unit $\varpi \in A^+$, and let $I$ denote the category of proper birational maps of schemes $f_i:X_i \to \mathrm{Spec}(A^+)$ which restrict to isomorphisms over the open subset $\mathrm{Spec}(A) \subset \mathrm{Spec}(A^+)$. This is cofiltered, because any $X_i \times_{\mathrm{Spec}(A^+)}X_{i'}\to \mathrm{Spec}(A^+)$ is still in $I$. Let $\overline{X_i} \subset X_i$ denote the vanishing loci of $\varpi$, so these are also cofiltered, and they all map compatibly to $\mathrm{Spec}(A^+/\varpi)$.

Now, there is a canonical map $\Phi: \mathrm{Spa}(A,A^+) \to \lim_{\leftarrow} \overline{X_i}$ given by the following recipe: for any point $x \in \mathrm{Spa}(A,A^+)$, let $K_x$ and $K_x^+$ be the associated residue field and valuation subring, respectively, so there are canonical maps $A \to K_x$ and $A^+ \to K_x^+$. We then get canonical compatible maps $\mathrm{Spec}(K_x) \to X_i$ for all $i$, simply because the loci in the $X_i$'s where $\varpi \neq 0$ all identify with $\mathrm{Spec}(A)$ by definition, so we can lift the given map $\mathrm{Spec}(K_x)\to \mathrm{Spec}(A)$ uniquely. But we also have a map $\mathrm{Spec}(K_x^+) \to \mathrm{Spec}(A^+)$, so by the valuative criterion of properness this lifts uniquely along each $f_i$, giving a compatible system of maps $r_i : \mathrm{Spec}(K_x^+) \to X_i$. Applying the $r_i$'s to the unique closed point of $\mathrm{Spec}(K_x^+)$ then gives the desired point in $\lim \overline{X_i}$.

Anyway, Bhatt proves that $\Phi$ is always a homeomorphism. The specialization map $\mathrm{sp}$ you want is just the composition of $\Phi$ with the natural map $\lim \overline{X_i} \to \mathrm{Spec}(A^+/\varpi)$. Of course, you don't need to go through all of this to *define* the specialization map, but it does seem to be the easiest way to prove the following:

**Fact:** The map $\mathrm{sp}$ is continuous, spectral, and closed.

Proof sketch: Continuity and spectrality follow from Bhatt's result together with some basic nonsense about limits of spectral spaces. Closedness can be deduced from the following useful criterion: any quasicompact, spectral, specializing map of locally spectral spaces is closed.