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.
Here are some sufficient conditions:
Let $A$ be a local domain, $K$ its fraction field, $L$ a finite extension of $K$, and $B$ the integral closure of $A$ in $L$. If $A$ is Noetherian and integrally closed, and $L$ is separable over $K$, then $B$ is necessarily finite over $A$, and so is semi-local (by going-up/down-type theorems).
If $A$ is not integrally closed, but its integral closure in $K$ is finite over itself, then again $B$ will be finite over $A$.
The condition that the integral closure of $A$ in $K$ be finite over $A$ is (at least by some people) called N-1. More generally, the condition that $B$ be finite over $A$ is called N-2, or Japanese. (The letter N here stands for Nagata, and I believe Grothendieck coined the adjecive Japanese for these rings because these properties were studied by Nagata and the commutative algebra school around him in Japan.)
So if $A$ is a Japanese ring, then $B$ will be finite over $A$, and hence semilocal. Of course, this is rather tautological: its utility follows from the fact that many rings (indeed, in some sense, most rings --- i.e. most of the rings that come up in algebraic number theory and algebraic geometry) are Japanese. E.g. all finitely generated algebras over a field, or over $\mathbb Z$, or over a complete local ring, are Japanese.
Here are some useful wikipedia entries related to this topic: Nagata rings and Excellent rings.
Best Answer
Let $L = K(\alpha)$ and let $f(x)$ be the minimal polynomial of $\alpha$.
Fact: there is a bijection between non-zero prime ideals of $A$ and factors of $f(x)$ in $K_P[x]$, where $P$ is the unique prime ideal of $K$.
Note that $K=K_P$ since $K$ is a complete discrete valuation field.
Now since we can assume that $f(x)$ is irreducible, there exists exactly one prime ideal $Q$ in $A$. Hence $A$ is a DVR.
Lastly, $L_Q = K_P \cdot L = K \cdot L = L$, as required.
As to your second question: $B$ being a finitely generated $A$-module does not require completeness. It is a standard result proved using the trace function.
Edit: $P = Q \cap R$. Showing that there is such a $Q$ can be done by appealing to the Lying Over Theorem. The fact that $A$ is a DVR, i.e. that there is only one prime ideal in $A$, is harder to prove: the only way I know of involves what I quoted as a "fact".
About $a_i$ lying in $A$:
We prove that for $x \in A$, $\operatorname{tr}_{L/K}(x) \in R$. Let $M$ be the Galois closure of $L/K$. Since $x$ is integral over $R$, $\sigma_i(x)$ is integral too for all $\sigma_i \in \operatorname{Gal}(M/K)$. Hence $\sum \sigma_i(x) = \operatorname{tr}_{L/K}(x)$ is integral over $R$. As a trace (fixed by the Galois action), this must lie in $K$. Therefore $\operatorname{tr}_{L/K}(x)$ is an element of $K$ integral over $R$, and as $R$ is integrally closed, $\operatorname{tr}_{L/K}(x) \in R$.