You give a method for producing a valuation, $\nu$, corresponding to a theory, $\Gamma$. This is well-defined, but $\nu$ might not make $\Gamma$ true! For instance: what happens if $\Gamma=\{\psi_0\iff\neg\psi_1\}$ for some atomic $\psi_0, \psi_1$? Then your valuation would make both $\psi_0$ and $\psi_1$ false, so $\Gamma$ would not hold!
Essentially, in the example above, we need to make a choice between making $\psi_0$ true and making $\psi_1$ true. The point of extending to a complete theory is exactly to make these sorts of choices.
In fact, if you prefer, we can phrase the argument in terms of making choices, instead of forming a completion. Given $\Gamma$, we list the atomic propositions as $\psi_0, \psi_1, . . .$; at stage $\alpha$, we have built a partial valuation $\nu_\alpha$ of the first $(\alpha-1)$-many atomic propositions. We then extent $\nu_\alpha$ to $\psi_\alpha$ as follows: $\nu_{\alpha+1}(\psi_\alpha)=1$ if for every finite subset $\Gamma_0$ of $\Gamma$, there is a valuation $\mu$ which makes $\Gamma_0$ true, extends $\nu_\alpha$ and such that $\mu(\psi_{\alpha})=1$; and we make $\nu_{\alpha+1}(\psi_\alpha)=0$ otherwise. If you unpack this, of course, this is really just forming a completion of $\Gamma$, but it might seem more intuitive.
It is true that, thanks to completeness (and soundness) theorem, proving Points 1 and 2 is equivalent to prove Points 3 and 4.
And your proof of Point 4 is correct.
But your proof of Point 3 is not correct, and actually Point 3 (and hence Point 1) does not hold!
Where is the error in your attempt to prove Point 3? From the fact that $\mathcal{A} \models \lnot \phi$ it does not follow that $\Sigma \models \lnot \phi$.
Indeed, $\Sigma \models \lnot \phi$ means that for every model $\mathcal{A}'$, if $\mathcal{A}' \models \Sigma$ then $\mathcal{A}' \models \lnot \phi$.
But in your attempt to prove Point 3, you have just shown that there exists a model $\mathcal{A}$ such that $\mathcal{A} \models \Sigma$ and $\mathcal{A} \models \lnot \phi$.
A priori, it is still possible that there is another model $\mathcal{B}\models \Sigma$ such that $\mathcal{B} \models \phi$, your argument does not exclude this possibility. And actually this is what actually happens!
Why Point 3 does not hold? Suppose that $\Sigma$ is the set of axioms defining a group (it can be easily expressed in first-order logic, see here), and that $\phi$ is the formula expressing that the group is abelian. Clearly, $\Sigma \not\models \phi$ because not all groups are abelian, but from that it does not follow that $\Sigma \models \lnot \phi$ because it is not true that all groups are non-abelian.
Best Answer
Suppose any consistent set of formulae is satisfiable. Suppose $\Gamma \not\vdash \phi$. Then by definition, $\Gamma \cup \{\neg\phi\}$ is consistent. By assumption, $\Gamma \cup \{\neg\phi\}$ has a model, so not every model of $\Gamma$ is a model of $\phi$, i.e. $\Gamma \not\vDash \phi$.
Suppose instead that $\Gamma \vDash \phi$ implies $\Gamma \vdash \phi$. Take any (non-empty) consistent set of sentences, say $\Sigma = \Gamma \cup \{\phi\}$. Since $\Gamma \cup \{\phi\}$ is consistent, $\Gamma \not\vdash \neg\phi$. By assumption, $\Gamma \not\vDash \neg\phi$, i.e. $\Gamma \cup \{\phi\}$ has a model. (Note, we can always write a non-empty consistent set this way, even if $\Gamma$ here is empty).