Why $\mathbb{Q}_p^{ur} \neq \widehat{\mathbb{Q}_p^{ur}}$? Why is $\mathbb{Q}_p^{ur}$ not complete?
And is there a criterion to know when $K^{ur} = \widehat{K^{ur}}$ ? (where $K$ is a p-adic field, i.e. a field of characteristic 0 that is complete with respect to a discrete valuation that has a perfect residue field of characteristic $p > 0$. ) 
 A: Even though you say that you’ve found an explanation in Gouvêa’s book that $\mathbb Q_p^{ur}$ is not complete, let me give one here, for completeness.
The resideu field $\mathbb F_p$ of $\mathbb Z_p\subset\mathbb Q_p$ has an infinite tower of extensions, $\mathbb F_p=k_0\subset k_1\subset k_2\subset\cdots$, eacb=h inclusion being proper. There are many such towers, of course. EAch of the fields $k_n$ is a simple extension of $\mathbb F_p$, say generated by $\alpha_n$, a primitive $s_n$-th root of unity of some order. Each of these $\alpha_n$’s may be lifted (uniquely) to a primitive $s_n$-th root of unity $\zeta_n$, necessarily in the maximal unramified extension of $\mathbb Q_p$. Now take the Cauchy series $\sum_np^n\zeta_n$, certainly in the completion, and not in any finite extension of $\mathbb Q_p$, thus not in the maximal unramified (since that’s an algebraic extension of $\mathbb Q_p$). [I’m confident that this argument is not as clear or as economical as Gouvêa’s.]
Now: take a complete valued field $K$ of characteristic zero whose residue field $\kappa$ has characteristic $p>0$. If $\kappa$ is algebraically closed, then $K$ is already its own maximal unramified. If $\kappa$ is not algebraically closed, however, then it has an infinite ascending chain of algebraic extensions just as in the paragraph above, and the rest of the proof there goes through for this situation, so that the maximal unramified of $K$ is not complete.
In other words, the only way for $K^{ur}$ to be complete is for $K$ to be equal to $K^{ur}$.
