I am trying to show the relationship between logical consequence of Gamma and being true in every model of Gamma A problem in my textbook is the following.
We say $\Gamma \models A  $ if the following holds: If $ I $ be any interpretation of $ L $ and $ \phi $ is any assignment that satisifies $ \Gamma ~~( \phi(B) = T \leftrightarrow B \in \Gamma),$ then $ \phi $ satisfies A.  
1) if $ \Gamma \models A $, then $ A $ is true in every model of $ \Gamma $. 
2) if every formula in $ \Gamma $ is a sentence, and if $ A $ is true in every model of $ \Gamma $, then $ \Gamma \models A $. 
3).the formula $ \forall x_1 Rx_1 $ is true in every model of $ \{ Rx_1 \} $,  yet $ \forall x_1 Rx_1 $ is not a logical consequence of $ Rx_1 $
$ ~~ $ 
For the first part what I said was, 
Assume $ \Gamma \models A $, yet there is model, M, of $ \Gamma $ such that $ A $ is not true. (1)
By definition of logical consequence, we have that for any $ \phi $ that satisfies $ \Gamma $ it follows that $ \phi(A) = T $.  (2)
As $ M $ is a model, we have that $ \phi(B) = T $ for all $ B \in \Gamma ~~$. (3)
$ \Rightarrow ~ \phi(A)  = T ~ in ~M~~~$ (4), from line 2 and 3 
$ \bot ~~  \Rightarrow  $ A is true in every model M.
$~$
Now, looking at the other question, my answer has completeny ignored the logical quantifiers, and whether A is a sentence of not didn't matter.  Is my proof for part 1) wrong? what am I missing? How should I proceed for the other parts? What does it really mean to be true in every model?
Thank you in advance.
 A: In this type of problems we have to use all the definition: what does it mean for a formula $A$ to be true in a model $M$ ? That for every $\phi$ we have that $M, \phi \vDash A$ ($\phi$ satisfies $A$ in $M$).
For 1), if $M$ is a model of $\Gamma$, this means that: for every $B \in \Gamma$ and every $\phi$, we have: $M,\phi \vDash B$.
But $\Gamma \vDash A$, i.e. $M,\phi \vDash A$, for every $\phi$. And this holds for every $M$ that is a model of $\Gamma$. 
Thus:

for every model of $\Gamma$ and every $\phi$ we have $M,\phi \vDash A$.

2) What about sentences ? 
Now the key property is that if $B$ is a sentence and $M$ is a model of $B$, then $M,\phi \vDash B$, fo revery $\phi$.
Let $M$ a model of $\Gamma$: this means that $M,\phi \vDash \Gamma$, for every $\phi$ (because all formulas in $\Gamma$ are sentences).
But $A$ is true in very model of $\Gamma$, i.e. $M,\phi \vDash A$, for very $\phi$ and every $M$ that is a model of $\Gamma$.
Thus:

$\Gamma \vDash A$.

3) is a counter-example showing that the proviso about $\Gamma$ (all formulas in $\Gamma$ are sentences) is necessary.
By 1) we have that $\forall x Rx$ is true in every model of $Rx$, because if $M$ is a model of $Rx$ this means that $M, \phi \vDash Rx$, for every $\phi$.
But thus also every $x$-variant of $\phi$ will satisfy $Rx$, and thus $M,\phi \vDash \forall xRx$.
Consider now a simple intepretation using $\mathbb N$ as domain and intepret $Rx$ as $(x=0)$.
Let $\phi$ such that $\phi(x)=0$; clearly $\mathbb N, \phi \vDash (x=0)$.
But $\mathbb N$ is not a model of $(x=0)$, because not every $\phi$ satisfies it.
And obviously $\forall x (x=0)$ is not true in $\mathbb N$.
Thus:

$(x=0) \nvDash \forall x (x=0)$.

