The base field of the Berkovich unit disk When defining the Berkovich unit disk $D_k$, we always start with the case that the base field $k$ is algebraically closed. This is because when k is not algebraically closed, we have the isomorphism $D_{\hat{k^a}}/Gal(K^a/K)\cong D_k$, where $k^a$ is the algebraic closure of k. My doubt is why do we need the condition of algebraic closeness? 
Basically, it's all about classifying the points and the tree structure. The definitions of Type I-IV of points in $D_k$ do not depend on the algebraic closeness of $k$ -- it's just descending chain of discs. Besides that, proving that every point $x$ in $D_k$, i.e. every multiplicative semi-norm on $k\langle T\rangle$, belongs to one of Type I-IV seems to be independent on the algebraic closeness of $k$ either. So what do I miss?
The following pictures are screenshots from Berkovich's book Spectral theory and analytic geometry over non-Archimedean fields:


The only suspicious thing that may break in the non-algebraically closed setting is the saying that

multiplicative seminorms on $k\{r^{-1}T\}$ are completely determined by their values on the polynomials of the type $T-a, a\in k$.

This is a corollary of the Weierstress preparation theorem. Do we need $k=k^{a}$ for that?
 A: I too find the statement saying that multiplicative seminorms are determined by the values on polynomials of the form $T-a$ suspicious. I suspect for the correct statement over a general field we need to consider all irreducible polynomials.
Although it's possible to work with a non-algebraically closed field directly, it's just more convenient to work over an algebraically closed field, because if you try to work with a non-algebraically closed field, you end up caring about elements in the algebraic closure and their Galois orbits anyway.
For example, consider type (I) points, then if $k \neq k^a$, you get type (I) points that don't correspond to points in $\{a \in k : |a|\leq 1\}$, because if $a \in k^a \setminus k$ with $|a|\leq 1$, then $f \mapsto |f(a)|$ corresponds to a point on $D_k$, even though $a \notin k$. So the algebraic closure shows up naturally. And then you need to worry about when two elements $a,b \in k^a$ with $a \neq b$ define the same type (I) point in $D_k$. Of course, we know that this happens exactly if $a$ and $b$ are in the same Galois orbit. But it seems easier to work first over the algebraic closure and then worry about the Galois action.
Also, this is a good example to test the proof against if we drop the algebraically-closed assumption: take $a \in k^a\setminus k$ with $|a|\leq 1$. For example, take $k=\Bbb Q_2$ and $a=\sqrt{2}$. Then for a polynomials of the form $f=T-b$ with $b \in \Bbb Q_2$, one computes that $|f(a)|=|a-b|=\max(|a|,|b|)=\max(\frac{1}{\sqrt{2}},|b|)$, because no element in $\Bbb Q_2$ has absolute value $\frac{1}{\sqrt{2}}$. But now if we had taken $a=\sqrt{6}$, then we would get the same result on polynomials of the form $T-b$ for $b \in \Bbb Q_p$! That is even though $\sqrt{2}$ is not Galois conjugate to $\sqrt{6}$ (as the only possibly Galois conjugates of $\sqrt{x}$ are $\pm \sqrt{x}$), so $\sqrt{2}$ and $\sqrt{6}$ define different points on $D_{\Bbb Q_2}$.
Similar remarks apply to the other types of points, e.g. for type (2) points, you can consider expansions around an element $a \in k^a \setminus k$ and you need to consider those if you want to classify all points.
