Proof of Upward Lowenheim-Skolem Theorem Currently I am reading Enderton's A Mathematical Introduction to Logic. 
In page $154,$ he stated the following: 

Lowenheim-Skolem Tarski Theorem Let $\Gamma$ be a set of formulas in a language of cardinality $\lambda,$ and assume that $\Gamma$ satisfies in some infinite structure.
  Then for every cardinal $\kappa\geq \lambda,$ there is a structure of cardinality $\kappa$ in which $\Gamma$ is satisfiable. 

The proof is as follows:

Proof: Let $\mathfrak{A}$ be the infinite structure in which $\Gamma$ is satisfiable. 
  Expand the language by adding a set $C$ of $\kappa$ new constant symbols.
  Let 
  $$\Sigma=\{c_1\neq c_2|c_1,c_2 \text{ distinct members of }C\}.$$
  Then every finite subset of $\Sigma\cup\Gamma$ is satisfiable in the structure $\mathfrak{A},$ expanded to assign distinct objects to the finitely many new constant symbols in the subset. (Since $\mathfrak{A}$ is infinite, there is room to accommodate any finite number of these.) So by Compactness $\Sigma\cup\Gamma$ is satisfiable, and by the Downward Lowenheim-Skolem Theorem it is satisfiable in a structure $\mathfrak{B}$ of cardinality $\leq \kappa.$ (The expanded language has cardinality $\lambda+\kappa=\kappa.$)
  But any model of $\Sigma$ clearly has cardinality $\geq \kappa.$
  So $\mathfrak{B}$ has cardinality $\kappa;$ restrict $\mathfrak{B}$ to the original language.

I have two questions: 

Questions 
  $(1):$ Why is every finite subset of $\Sigma\cup\Gamma$ satisfiable in the structure $\mathfrak{A}?$ If the finite subset is in $\Gamma$ only, then by assumption, the finite subset is satisfiable in $\mathfrak{A}.$ 
  However, when the finite subset contains some elements from $\Sigma,$ I fail to understand why it is satisfiable in $\mathfrak{A}.$
$(2):$ Why is every model of $\Sigma$ has cardinality $\geq \kappa?$

Any hint would be appreciated. 
 A: We start from a theory $\Gamma$ in a language $\mathcal L$.
Then we expand the language adding infinite many new individual constant $c_i \in C$, where $C$ has cardinalty $\kappa$.
Then we consider the set of sentence $\Sigma = \{ c_1 \ne c_2 \mid c_1, c2 \text {  distinct elements of   } C \}$.
Then we consider the set of sentences: $\Gamma \cup \Sigma$ in the expanded language.
Lastly, we assume that $\Gamma$ has an infinite model $\mathfrak A$.
(2) Due to the fact that the expanded language has $\kappa$ many distinct constants, the "axiom" $c_i \ne c_j$ of $Σ$ forces us to interpret them with distict objects. 
And (1) due to the fact that $\mathfrak A$ has an infinite domain, it has enough objects to satisfy a finite number of new "axioms" from $\Sigma$.
Thus, by compactness, there is model of $Γ \cup Σ$ and by Downward L-S there is a model $\mathfrak B$ (different from $\mathfrak A$) with the required cardinality. 
Obviously, being a model of $Γ \cup Σ$, $\mathfrak B$ is also a model of $Γ$.
