A paper I'm reading implicitly assumes the statement: Let $K_0$ be the completion of $\mathbb {Q}_ p^{un}$. Then any finite extension of $K_0$ is complete with residue field $\bar {\mathbb {F}} _p$. So here are a few questions:
1. Why is this true? Is it true in general that if you complete, and then take algebraic closure - then that is complete?
2. Is it true that any finite extension of $K _0$ comes from the completion of a finite extension of $\mathbb {Q} _p ^ {un}$?
A finite extension of a complete field is complete. This is standard and proved in lots of places e.g. one of the first two chapters of Cassels-Froehlich, or Bosch-Guentzer-Remmert, or lots of other places. The residue field will only go up a finite amount too, so it must have stayed the same!
Your Q1 appears to have two parts: the first is "why is this true", which I think I just answered. The second part is not true: if you complete Q at p you get Q_p and if you then take alg closure you get Q_p-bar which isn't complete. What is true is that if you take alg closure and then complete, you're still alg closed. This is proved in BGR too.
Q2 is also true and this will follow from Krasner's lemma, also proved in BGR (and in lots of other places). A finite extension of K_0 is K_0(alpha) with alpha a root of a polynomial in K_0[x]. By Krasner I can perturb the coefficients of this polynomial without changing the extension. So I perturb them until they're in Q_p^{un} and there's my finite extension of Q_p^{un}.
1) The field K_0 is the fraction field of the Witt vectors of F_p-bar (my non-TeX notation for the algebraic closure of F_p). Call that ring W. It is a complete DVR with an alg. closed residue field F_p-bar.
Any finite extension field of K_0 is naturally a finite-dimensional K_0-vector space. Since K_0 is complete, all vector space norms on a fin. dim. K_0-vector space V (say, the sup-norm w.r.t. some basis, but truly any vector space norm can be used) are equivalent to each other: each is bounded above and below by a constant multiple of any other norm. So all these norms provide V with the same topology. It's just the product topology on a direct sum of copies of K_0 (as many as the dimension), i.e. the topology of componentwise convergence if you pick a basis and look at the norm attached to that choice of basis.
Since the abs. value on K_0 is complete, it has a unique extension to an abs. value on any finite extension field K. That extended absolute value on K is a special type of K_0-vector space norm (namely it is multiplicative, going beyond the strict conditions of a vector space norm), but it's a vector space norm all the same, so the topology it puts on K is just coefficientwise convergence in a basis. Thus visibly it is complete. This all has nothing to do with K containing a completion of Q_p-unram.
The residue field of a finite extension of a complete valued field is a finite extension of the residue field of the initial field. So if the bottom residue field is F_p-bar, so alg. closed, then certainly any finite extension of your field will still have residue field F_p-bar. It's similar to why residue fields at points on Riemann surfaces are all C, since C is alg. closed.
To address the second part of question 1, completing and taking alg. closure doesn't produce a complete field in general. Try passing from Q to Q_p to its alg. closure.
2) Since the Witt vectors W = W(F_p-bar) is/are a complete DVR with alg. closed residue field, the finite extensions of its fraction field involve no unram. extensions (the res. field can't grow anymore). All finite extensions are totally ramified, so generated by the root of an Eisenstein polynomial with coeff. in W. By Krasner's lemma, you can get the same extension by adjoining a root of a poly. with coeff. close to those of your original polynomial. Therefore by choosing nearby coeff. in W(F_q) for large q, you get an Eisenstein poly. with coeff. in Q_p-unram, and adding a root of that to Q_p-unram gives you a finite extension of Q_-unram which is dense in your finite extension of K_0.
