Simplest example of failure of finite Galois descent in algebraic $K$-theory? Let $E \to F$ be a $G$-Galois extension of fields. 
What is the simplest example where the natural map $K(E) \to K(F)^{hG}$ is not an equivalence on connective covers (i.e., where finite Galois descent fails)? 
 A: First of all, by the long exact sequence in homotopy, it suffices to give a $G$-Galois extension $E \subset E'$ such that the homotopy fiber of $K(E) \to K(E')^{hG}$ contains a class in degree $-1$. This means that the map $\pi_0 K(E) \to \pi_0 K(E')^{hG}$ is not surjective. 
The spectrum $\mathrm{fib}( K(E) \to K(E')^{hG} )$ is annihilated by the group order $|G|$. This follows from a classical argument with the transfer: anything in the image of the transfer $K(E')_{hG} \to K(E)$ annihilates the homotopy fiber. (In the language of $G$-spectra: if $R$ is a $G$-ring spectrum, then any class in $\pi_0 R^G$ which is in the image of $\pi_0 R_{hG} \to \pi_0 R^G$ annihilates the fiber of $R^G \to R^{hG}$). Note that the order of $G$ is the image of $1$ under the transfer map for $K$-theory. It follows that $\mathrm{fib}( K(E) \to K(E')^{hG} )$ splits as a product of its $l$-adic completions for various primes $l$. It thus suffices to show that $\pi_{-1}$ of some $l$-adic completion is nonzero, or equivalently that $l$-adically completed $K$-theory is not a sheaf of connective spectra. 
If I'm not mistaken, the following (based on an idea of Dustin Clausen) should give an example. Let $E$ be a field of characteristic $\neq l$ with the following two properties:
1) $E$ contains all the $l$-th power roots of unity
2) $H^2( \mathrm{Gal}(E); \mathbb{Z}_l(1)) \neq 0$.
Then I claim the functor $E' \mapsto K(E')$ is not a sheaf of connective spectra on the etale site of $E'$ (equivalently doesn't satisfy Galois descent). It suffices  to work after $l$-adic completion, and for the rest of this post I'll write $\hat{K}$ for the $l$-adic completion of $K$. 
In fact, I claim that there exists a finite Galois extension $E'$ of $E$ such that $\hat{K}(E')^{hG}$ is bigger than $\mathbb{Z}_l$. 
To see this, choose a finite Galois extension $E'$ of $E$ with group $G$ such that $H^2(G; \mathbb{Z}_l) \neq 0$. I claim that $\hat{K}(E')^{hG}$ has a $\pi_0$ larger than $\mathbb{Z}$. To see this, we observe that there is a map $\mathbb{Z}_l \to \pi_2 ( \hat{K}(E)) \to  \pi_2 (\hat{K}(E'))$ (recall that we're $l$-completing everywhere) and the composite to $\mathbb{Z}_l \to \pi_2( \hat{K}(E')) \to \pi_2( \hat{K}(E^{bar}))$ is an isomorphism (the latter by Suslin).
By our assumptions, we thus get a cycle in $H^2( G, \pi_2 \hat{K}(E'))$ which maps to a nonzero cycle in $H^2(E, \mathbb{Z}_l(1))$.
To see that it survives the HFPSS, it suffices to work at the level of spaces. But there is a $G$-equivariant map (where the hat denotes completion) $\widehat{BGL_1(E')} \to \Omega^\infty \hat{K}(E')$. On $\pi_2$, we get exactly a $\mathbb{Z}_l$ hitting the class desired. Since $BGL_1(E')$ has no higher homotopy, we find that the class in $H^2( G, \pi_2 \hat{K}(E'))$ must survive as it comes from $(\widehat{BGL_1(E')})^{hG}$.
Since this permanent cycle is detected nontrivially in the descent ss for the full Galois group, it must be nonzero. Thus, $\pi_0 \hat{K}(E')^{hG}$ is bigger than $\mathbb{Z}_l$.
