TITLE: In model theory, does compactness easily imply completeness?
QUESTION [22 upvotes]: Recall the two following fundamental theorems of mathematical logic:
Completeness Theorem: A theory T is syntactically consistent -- i.e., for no statement P can the statement "P and (not P)" be formally deduced from T -- if and only if it is semantically consistent: i.e., there exists a model of T.
Compactness Theorem: A theory T is semantically consistent iff every finite subset of T is semantically consistent.  
It is well-known that the Compactness Theorem is an almost immediate consequence of the 
Completeness Theorem: assuming completeness, if T is inconsistent, then one can deduce 
"P and (not P)" in a finite number of steps, hence using only finitely many sentences of T.
The traditional proof of the completeness theorem is rather long and tedious: for instance, the book Models and Ultraproducts by Bell and Slomson takes two chapters to establish it, and Marker's Model Theory: An Introduction omits the proof entirely.  There is a quicker proof due to Henkin (it appears e.g. on Terry Tao's blog), but it is still relatively involved.  
On the other hand, there is a short and elegant proof of the compactness theorem using ultraproducts (again given in Bell and Slomson).  
So I wonder: can one deduce completeness from compactness by some argument which is easier than Henkin's proof of completeness?  
As a remark, I believe that these two theorems are equivalent in a formal sense: i.e., they are each equivalent in ZF to the Boolean Prime Ideal Theorem.  I am asking about a more informal notion of equivalence.  

UPDATE: I accepted Joel David Hamkins' answer because it was interesting and informative.  Nevertheless, I remain open to the possibility that (some reasonable particular version of) the completeness theorem can be easily deduced from compactness.

REPLY [7 votes]: About the equivalence of the compactness, completeness, prime ideal theorems over ZF: what really matters here is the case when the language $L$ over which the theory $T$ is defined is not well-ordered. Otherwise, the Henkin proof gives a model of $T$ without using any form of axiom of choice. In particular, it is OK when the considered language is countable.
Now, the implication compactness $\Rightarrow$ completeness in the general case goes as follows (although it still uses completeness for well-ordered theories, which is a theorem of ZF).
Fix a first-order language $L$. Let $T$ be a syntactically consistent theory. Then any finite $F\subseteq T$ is syntactically consistent. Define $L_F$ to be the language whose operational and relational symbols are the ones occurring in $F$. Since $F$ is finite, $L_F$ is finite. Then $F$ is a syntactically consistent theory in the language $L_F$. We have completeness for countable languages, so we have a model $M_F$ of $F$ treated as a theory over $L_F$. The model $M_F$ can easily be extended to a model $M_F'$ of $F$ treated as a theory over $L$ (just give the unused symbols trivial interpretations: empty relations and constant operations). By compactness we now have a model of $T$.