I am trying to learn about $p$-adic numbers using Robert's A Course in $p$-adic Analysis. In section II.1.4 he proves that the algebraic closure,$\mathbb{Q}_p^a$, of $\mathbb{Q}_p$ is not complete by showing it's not a Baire space. The proof goes as follows:
Let us define the sequence of subsets $$X_n= \{x\in\mathbb{Q}_p^a\mid [\mathbb{Q}_p(x):\mathbb{Q}_p]=n\},\quad n\geq 1\quad(1)$$ so that $\mathbb{Q}_p^a=\bigcup X_n$. It is also obvious that $\lambda X_n\subset X_n$ ($\lambda\in\mathbb{Q}_p$), $X_m+X_n\subset X_{mn}$, and in particular $$X_n+X_n\subset X_{n^2}.\quad (2)$$
$\quad$ (a) These subsets are closed. If $x\neq 0$ is in the closure of $X_n$, say $x=\lim x_i$ with a sequence $(x_i)$ in $X_n$, then for each $x_i$, let $f_i(X)\in\mathbb{Q}_p[X]$ be a polynomial of least degree with $x_i$ as a root and coefficients scaled so they lie in $\mathbb{Z}$ and at least one of them is in $\mathbb{Z}_p^\times$. Extracting if necessary a subsequence of $(f_i)$, we can assume that it converges (in norm, coefficient-wise), say $f_i\to f$, so $f\in\mathbb{Z}_p[X]$ has degree less than or equal to $n$ and at least one coefficient in $\mathbb{Z}_p^\times$, so $f(X)\neq 0$. By the ultrametric property, the convergence $f_i\to f$ is uniform on all bounded sets of $\mathbb{Q}_p^a$. Since the convergent sequence $(x_i)$ is bounded, we have $$f(x)-f_i(x_i)=\underbrace{f(x)-f(x_i)}_{\to 0}+\underbrace{f(x_i)-f_i(x_i)}_{\to 0}\to 0.$$ This implies $f(x)=\lim f_i(x_i)=0$ and $x\in X_n.\,\,(3)$
$\quad$ (b) The subset $X_n$ have no interior point. Since for any closed ball $B$ of positive radius in $\mathbb{Q}_p^a$ we have $\mathbb{Q}_p^a=\mathbb{Q}_p\cdot B$ $(4)$, such ball cannot be contained in a subset $X_n$, and no translates can be contained in $X_n$.
It seems like the definition in $(1)$ should be $X_n =\{x\in\mathbb{Q}_p^a\mid [\mathbb{Q}_p(x):\mathbb{Q}_p]\leq n\}$ so the equality in $(2)$ holds. Otherwise for $x\in X_n$, $2x\in X_n+X_n$ but is not in $X_{n^2}$ since the degree of $x$ and $2x$ will be the same. Moreover, in $(3)$ Robert claims $x\in X_n$ since $f(x)=0$ and $\deg f\leq n$, which only works if $X_n$ is the set of all elements of degree at most $n$ over $\mathbb{Q}_p$. The problem with defining $X_n$ in this way is that it no longer has an empty interior since $\mathbb{Q}_p\subset X_n$. So I am not sure what am I missing here.
I'm also unsure why the equality in $(4)$ holds ($\mathbb{Q}_p^a=\mathbb{Q}_p\cdot B$).
Any help is appreciated!
Indeed it also seems to me that the definition of $X_n$ should be modified as
$$ X_n= \{x\in\mathbb{Q}^a_p \mid [\mathbb{Q}_p(x):\mathbb{Q}_p] \leq n\}, \qquad n\geq 1. $$
Also, using this modified definition, $X_n$ is still nowhere dense since it cannot contain any ball of positive radius. I think you confused yourself by believing that $\mathbb{Q}_p = X_1$ is open in $\mathbb{Q}^a_p$, which is not true. (Although not a good analogy, think of the fact that $\mathbb{R}$ is not an open subset of $\mathbb{C}=\mathbb{R}^a$.)
As for why $\mathbb{Q}^a_p=\mathbb{Q}_p\cdot B$, I think the proof is referring to a closed ball $B$ centered at $0$. (In that case, the identity is almost obvious.) Then it also makes sense that the proof mentions "translates", i.e., close balls not necessarily centered at $0$.