Extension of places to an algebraic closure Let $K$ be a number field and $v$ a place of $K$. One often extends $v$ to a place $\overline{v}$ of an algebraic closure $\overline{K}$ of $K$, which is equivalent to choosing an embedding $\overline{K} \hookrightarrow \overline{K_v}$, where $\overline{K_v}$ denotes an algebraic closure of the completion $K_v$. However, I haven't found an explanation of how this extension process is done concretely. Can you help me on this one? 

The method suggested by Brandon Carter in his comments will surely work, but I think an explicit call to the Axiom of Choice will be necessary to make the process succeed.
You seem to recognize that there are two ways of attacking the question: by finding an extension of a $K$-place to $\overline K$, and embedding $\overline K$ into an algebraic closure of $K_v$, the completion of $K$ with respect to $v$. In my opinion, whichever approach you use, you’ll need AC.
I had rather use the second approach: from $K\hookrightarrow\overline{K_v}$ and $K\hookrightarrow\overline K$ you conclude that there is a map $\overline K\to\overline{K_v}$ making the triangle commutative. It’s a standard fact that a morphism $\phi$ from $K$ into an algebraically closed field $\Omega$ (not necessarily an algebraic closure of $K$) may be extended to any algebraic extension $K'$, that is, there is (in fact are many) $\phi':K'\to\Omega$.
You prove this standard fact by setting up the partially ordered set of all $(L,\psi)$ for which $K\subset L\subset K'$ and $\psi$ extends the given $\phi$. The ordering is the natural one of extension, and you just do a standard Zorn’s Lemma argument on it. You get a maximal member $(\Lambda,\Psi)$, and you then do the inductive step of arguing that $\Lambda$ must equal $K'$.
If this was too sketchy, don’t hesitate to e-mail me for details.
