Model with constant symbols as domain Work in a countable language $\mathcal{L}$ containing constants $c_n$ for all $n \geq 1$. Let $T$ be a satisfiable set of $\mathcal{L}$-sentences such that: if $\phi(v)$ is a formula with $T \vDash \phi(c_n)$ for all $n$, then $T \vDash \forall v \phi(v)$.
Show that $T$ has a model $\mathcal{A}$ with domain $A = \{ c_n^\mathcal{A} : n \geq 1\}$.
My thought is that we want to find some way to mimic the Henkin proof by creating a term model. This process itself though introduces new constants, so would not work in this case. Thanks for any help.
 A: Hint: you want a model that omits the type $\{x \neq c_n : n \ge 1\}$.
A: EDIT: I think I may have misread the question - the following assumes that the language consists only of constant symbols.

Levon suggests using the omitting types theorem; this will work, but there's another more elementary approach.
I'll first prove a seemingly weaker fact, then show how it is actually the meat of the whole problem.

Claim: if $T\models c_i\not=c_j$ for each $i\not=j$, then $T$ has a model consisting entirely of constants.

Proof. By compactness + Lowenheim-Skolem, we have a countable model $M$ of $T$ with infinitely many "unnamed elements" (= elements not the interpretation of any constant symbol). For fixed $n$, let $L_n$ be $\{c_1, c_2, ..., c_n\}$ and let $T_n$ be the set of $L_n$-consequences of $T$. Finally, let $M_n$ be the reduct of $M$ to $L_n$ (so forget about all the remaining constants).
Now consider the substructure $N\subset M$ consisting only of the named elements, that is, $N=\{c_i^M: i\in\mathbb{N}\}$. Define $N_n$ similarly to $M_n$ above. The crucial observation is the following:

For each $n$, we have $M_n\cong N_n$.

This is because in each structure we have finitely many named elements and infinitely many unnamed ones; note that by contrast $M\not\cong N$. 
So what? Well, fix $\varphi\in T$; I'll show $N\models\varphi$. Each $\varphi\in T$ uses only finitely many constant symbols - that is, each $\varphi\in T$ is in some $T_n$ (put another way, $T=\bigcup_{n\in\mathbb{N}} T_n$). Reducts don't change satisfaction, so $M\models \varphi$ iff $M_n\models \varphi$; similarly, isomorphisms don't change satisfaction, so $N_n\models\varphi$; and expansions don't change satisfaction, so $N\models\varphi$.
So $N\models T$, and consists only of constant symbols. $\quad\Box$
Note that this argument in fact shows that $N\prec M$.

Now on to the whole problem. First, by compactness we may assume that for each $i, j$, either $T\models c_i=c_j$ or $T\models c_i\not=c_j$ by expanding $T$ if necessary while preserving satisfiability. Call this assumption "$(*)$", to distinguish it from the assumption in your post, which I'll denote "$(+)$".
We first rule out the case where $T$ has a finite model. Note that above, we never used $(+)$! We actually only need it for this step:

Claim: if $T$ has a finite model, then $T$ in fact only has models where every element is named by a constant.

Proof. Suppose $T$ has a finite model. Then by $(*)$, for some finite set $\{i_1, ..., i_n\}$ we have for each $k$ that $T\models \bigvee_{1\le j\le n}c_k=c_{i_j}$; hence by $(+)$, $T$ satisfies $\forall x \bigvee_{1\le j\le n}x=c_{i_j}$, and so only has models consisting only of constants. And in fact it's not hard to show that $T$ has only one model after all.) $\quad\Box$
So now we're just left with the case where $T$ has only infinite models. In this case, there must be an infinite set $S\subseteq\mathbb{N}$ such that $T\models c_s\not=c_t$ for distinct $s, t\in S$ (why?), and so without loss of generality we may in fact assume that $T\models c_i\not=c_j$ for each distinct $i, j\in\mathbb{N}$ (why?) - bringing us right back to the claim at the beginning of the answer. 
