Let $K$ be a finite extension of $\mathbf{Q}_p$, i.e., a $p$-adic field. (Is this standard terminology?)
Why is (or why isn't) an algebraic closure $\overline{K}$ complete?
Maybe this holds more generally:
Let $K$ be a complete Hausdorff discrete valuation field. Then, why is $\overline{K}$ complete?
I think I can show that finite extensions of complete discrete valuation fields are complete.
Best Answer
The algebraic closure has countably infinite dimension over $\mathbb Q_p$, and therefore (by the Baire category theorem) is not metrically complete. (Except the case $\mathbb Q_\infty = \mathbb R$, where the algebraic closure has finite dimension, and is metrically complete.)
explanation
In $\mathbb Q_p$, let $x_n$ be a solution of $X^{n}=p$. Then $\{x_2,x_3,x_4,\dots\}$ is linearly independent over $\mathbb Q_p$. But we still need a proof that the algebraic closure does not have uncountable dimension.
Torsten Schoenberg provided the missing part:
Countable dimension follows from Krasner / Hensel and compactness of $\mathbb Z_p$ which shows for each $d \in \mathbb N$ , $\mathbb Q_p$ only has only countably many (actually finitely many) extension of degree $d$ in $\overline{\mathbb Q_p}$. I doubt there is a more elementary argument for that.
former explanation
We see in Incompleteness of algebraic closure of p-adic number field, from $9$ years ago question that it is incorrect...
How about an example? In $\mathbb Q_2$, the partial sums of the series $$ \sum_{n=1}^\infty 2^{n+1/n} $$ belong to $\overline{\mathbb Q_2}$, but the sum of the series does not. The partial sums form a Cauchy sequence with no limit in $\overline{\mathbb Q_2}$.
Why does this sum not exist in $\overline{\mathbb Q_2}$ ?
It is not trivial, but interesting: any $x$ which is algebraic of degree $n$ over $\mathbb Q_2$ has a unique series expansion $$ x = \sum 2^{u_j} $$ where $u_j \to \infty$ (unless it is a finite sum) and all $u_j$ are rationals with denominator that divides $n!$. (Maybe divides $n$ in fact?) But the series expansion in this example has arbitrarily large denominators.