Why doesn't "$V$ is a model of $ZF$" imply consistency of ZF? One can prove that $V$, Von Neumann's universe, satisfies all of $ZF$ axioms, so it's a model of $ZF$. But I can't see why this doesn't imply the consistency of that theorem. I'm aware that consistency would follow if model was a set, but why doesn't it work in case of proper classes?
I mean, $V$ looks to be perfectly well defined and well behaved structure, in which e.g. empty set isn't non-empty, but if $ZF$ was inconsistent, then it would (of course) prove that empty set is non-empty. Where is the flaw in this?
 A: You seem to already have answered your question when you realized that in order to deduce the sentence $Con(ZFC)$ using Gödel's completeness theorem, you need to have a set model existence of which is impossible to be proven by the incompleteness theorem.
If you believe that there exists a Platonic universe of sets where certain objects exist and satisfy axioms of ZFC, then you believe that ZFC is consistent because there is a "model" of it. On the other hand, if you try to formalize this belief within ZFC and want to prove that ZFC is (formally) consistent, then you need to have a set model.
Let me give another example which may clarify things for you. As you might already know, it is a theorem that L is a model of ZFC. But what does this theorem really mean, how do we express it in ZFC? Since L is a proper class, you cannot use the formalized truth predicate (which is defined only for sets) to say each axiom of ZFC is true in L.
Instead, what you do when you want to claim that a definable proper class M is a model of ZFC is that you relativize your quantifiers to this proper class and show that each axiom of ZFC is true, i.e. you want to prove $\phi^M$ for all axioms $\phi$.
From this perspective, what does it mean to say "V is a model of ZFC"? V is the universe, which you can define by $\phi(x): x=x$. Then the equivalence $\phi^V \leftrightarrow \phi$ is trivially true for all sentences, right?
The issue is expressing "is a model of ZFC" part in a sensible way for proper classes. As Joel suggested, you can work with stronger set theories.
A: The main problem with your proposal is that in ZFC we aren't actually able to express the statement "every axiom of ZFC is true in $V$". Expressing this statement makes use of a Tarskian truth predicate for first-order truth, and in ZFC, we can prove by Tarski's theorem on the non-definability of truth that there is no such definable predicate. So in a sense, the proposal doesn't even get off the ground.
When you say that we can prove that $V$ satisfies all the ZF axioms, this is not a single statement, but rather a scheme of infinitely many statements. That is, what we have are the separate facts, about each particular axiom $\sigma$ of ZFC in the meta-theory, that $\sigma$ is true in $V$. But we do not have the universal statement that would tie them together, "every axiom $\sigma$ of ZFC is true in $V$," and indeed this universal statement is not first-order expressible.
Meanwhile, your proof idea is actually resurrected in stronger theories that are able to prove that there is a truth predicate for first-order truth. I recently wrote a blog post, Kelley-Morse set theory implies Con(ZFC) and much more, for example, in which I explain how to prove the existence of a truth-predicate for first-order truth in Kelly-Morse set theory. Every axiom of ZFC comes out as true in $V$ according to this predicate, and this is formalizable in the stronger context of KM, and one can use this to prove that KM implies Con(ZFC) and much more, following essentially just the line of argument to which you allude. 
