Field of algebraic numbers over Q with p-adic value 
Define $\overline{\mathbb{Q}} \subset \mathbb{C}$ to be the subset consisting of all complex numbers which are algebraic over $\mathbb{Q}$. We know that $\overline{\mathbb{Q}}$ is a countable field and that is algebraically closed.

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*Show that there exists a sequence of finite extensions $E_{0}=Q  \subset E_{1} \subset \ldots \subset E_{n} \subset \ldots \overline{\mathbb{Q}}$, i.e. each $E_{i}/E_{i-1}$ is a finite exntesion and $\overline{\mathbb{Q}} = \cup_{n} E_{n}$.

*(Using the above) show that for any prime $p$, the $p$-adic absolute value extends to an absolute value on $\overline{\mathbb{Q}}$.


So, I proved 1 (just define $E_{i} = \mathbb{Q} (a_{1}, a_{2},\ldots, a_{i}$, where $\overline{\mathbb{Q}} = \left\{a_{1}, a_{2},\ldots\right\}$ ), but I don't know how to formalize $2$. Of course, using the fact that every nonarchimedean absolute value on a field, extends in at least one way to every finite extension, we get an extension of the p-adic valuation on each $E_{n}$, but I don't see how to end $2$. Perhaps, a continuity argument?
Thanks.
 A: As you say, you get an extension of the $p$-adic valuation to each $E_n$, and every element of Q-bar is in $E_n$ for some $n$, so you have extended the $p$-adic valuation to Q-bar, haven't you? 
A: Note that the extension of the $p$-adic valuation to $E_n$ is typically not unique; the possible extensions are in bijection with the number of primes in the ring of integers of $E_n$ lying over the prime $p$.
So for each $n$ there is a (non-empty!) finite set $S_n$ of extensions of the $p$-adic valuation to $E_n$.  Restricting an extension from $E_{n+1}$ to $E_n$ gives a map $S_{n+1} \to S_n$.  So you have a projective sequence of finite sets
$$ \cdots \to S_{n+1} \to S_n \to \cdots \to S_1 \to S_0$$
and you are trying to choose an element from each in a compatible fashion, i.e. you are trying to choose an element in the projective limit.  This will be possible if (and only if!) the projective limit is non-empty.
Added: As Gerry points out in (the comments to) his answer, these maps are 
surjective, and so one can just successively choose an element of each $S_n$
that maps to the previous choice in $S_{n-1}$; this gives the desired extensino.
Earlier unnecessary discussion:
General fact: the projective limit of a projective sequence of non-empty finite sets is always non-empty.  
Proof: A special case of a more general fact about projective limits called the Mittag--Leffler property, which is also pretty easy to prove directly (although not completely trivial if you've never thought about this kind of things before).  
Added: When the maps $S_{n+1} \to S_n$ are surjective (as they are in our context), this really
is trivial; as was already noted above in this special case, just choose a point in each $S_{n+1}$ mapping to the previous choice in $S_n$!
