If $T$ is $\kappa$-stable, then $|T|\leqslant\kappa$ I have heard that if $T$ is a $\kappa$-stable theory (say over a language $\mathcal{L}$), then there is an $\mathcal{L}$-theory $T'$ such that $|T'|\leqslant\kappa$ and $T'\models T$. I think I have a proof of this as follows:
Denote the constant symbols of $\mathcal{L}$ by $C$. First note that we can assume distinct symbols of $C$ are distinct modulo $T$. (Indeed, given any $c,d\in C$ distinct constant symbols, if $T\models c=d$ then we can replace every instance of $d$ in a sentence of $T$ by $c$ and so have that $T\upharpoonright_{\mathcal{L}\setminus\{d\}}\models T$.)
Now note that we may suppose $|C|\leqslant\kappa$. Indeed, for any $c\neq d\in C$, by the above we have that $\text{tp}(c)\neq\text{tp}(d)\in \text{S}_1(T)$, and so $|C|>\kappa$ would contradict $\kappa$-stability of $T$.
Indeed, because $T$ is $\kappa$-stable, we have that $|\text{S}_1(T)|\leqslant\kappa$. For any $p\neq q\in\text{S}_1(T)$, choose $\varphi_{p,q}(v)$ an $\mathcal{L}$-formula such that $\varphi_{p,q}\in p$ and $\neg\varphi_{p,q}\in q$. We claim that any $\mathcal{L}$-formula $\psi(v)$ consistent with $T$ is equivalent to a boolean combination of finitely many of the $\varphi_{p,q}(v)$.
Indeed, for any $r(v)\in\text{S}_1(T)$ such that $\psi\in r$, let $\Sigma_r(v)=\{\varphi_{r,q}\}_{q\neq r\in\text{S}_1(T)}$. We claim $\Sigma_r(v)\models\psi(v)$. If not, then $\Sigma_r(v)\cup\{\neg\psi(v)\}$ is consistent and so is contained in some complete type $s\in\text{S}_1(T)$. Note that $r\neq s$ as the former contains $\psi$ and the latter contains $\neg\psi$, and so we have $\varphi_{r,s}\in\Sigma_r(v)$. However, by construction $\Sigma_r(v)\subseteq s$, so this would mean $\varphi_{r,s}\in s$, a contradiction. Thus indeed $\Sigma_r(v)\models\psi(v)$ and so by compactness there is a finite subset $\Delta_r(v)\subseteq\Sigma_r(v)$ such that $\Delta_r(v)\models\psi(v)$. For each $r\ni\psi$, let $\theta_r(v)=\bigwedge\Delta_r$, so that $\theta_r\in r$, $\theta_r\models\psi$, and $\theta_r$ is a conjunction of (finitely many of) the $\varphi_{r,q}$.
Now we claim that $\{\neg\theta_r\}_{r\ni\psi}\models\neg\psi$. Indeed, if not, then $\{\neg\theta_r\}_{r\ni\psi}\cup\{\psi\}$ would be consistent and hence contained in a complete type $t\in\text{S}_1(T)$. But then $t\ni\psi$, so we would have $\theta_t\in t$ and $\neg\theta_t\in t$, a contradiction. Thus indeed $\{\neg\theta_r\}_{r\ni\psi}\models\neg\psi$, and so again by compactness there are $r_1,\dots,r_k\in \text{S}_1(T)$ such that $\{\neg\theta_{r_i}\}_{i=1}^k\models\neg\psi$. Let $\theta=\bigvee_{i=1}^k\theta_{r_i}$. Then $\neg\theta\models\neg\psi$, so by contraposition $\psi\models\theta$, and also each $\theta_{r_i}\models\psi$, so $\theta\models\psi$, and so $\psi$ is logically equivalent to $\theta$. In addition, each $\theta_{r_i}$ is a conjunction of finitely many $\varphi_{r,q}$, and so $\theta$ is a boolean combination of (finitely many of) the $\varphi_{p,q}$, as desired.
Now, $T\subset\{\psi(c):c\in C,\psi\text{ an }\mathcal{L}\text{-formula}\}$. Hence, letting $T'=\{\varphi_{p,q}(c):c\in C,p\neq q\in\text{S}_1(T), T\models\varphi_{p,q}(c)\}$, the above argument shows that $T'\models T$. Also, $|C|\leqslant\kappa$, and – since $|\text{S}_1(T)|\leqslant\kappa$ – also $|\{\varphi_{p,q}\}_{p\neq q\in\text{S}_1(T)}|\leqslant\kappa$, so $|T'|\leqslant\kappa$ as desired.
So, I have two questions – first, is this proof correct? Second, is there a quicker way of seeing this result? The condition that $|\text{S}_1(T)|\leqslant\kappa$ seems to me to be much weaker that $\kappa$-stability, so I feel like there should be a more immediate way of seeing this result that doesn't involve fiddling around with constant symbols, etc. Can we use the strength of full $\kappa$-stability somehow?
Edit: I think I have found a hole in my argument; in paragraph 5, I think the fact that $\Sigma_r(v)\cup\{\neg\psi(v)\}$ is consistent does not necessarily mean it is contained in an element of $\text{S}_1(T)$, for which we need also the condition that $\Sigma(v)\cup\{\neg\psi(v)\}$ is consistent with $T$. (For instance, if $\psi(v)$ is actually a sentence (so that the variable $v$ does not appear (free) in it) then $\neg\psi$ will never be consistent with $T$ by the assumption that $T$ is complete and that $\psi$ is consistent with $T$.) Is there a way to fix this hole? Note that the cases where $\psi(v)$ is a sentence are actually very important, as they are the only way to account for those sentences of $T$ that contain no constant symbols.
 A: Unfortunately, the thing you're trying to prove is not true (althought the slogan "If $T$ is $\kappa$-stable, then $|T|\leq \kappa$" is essentially correct - see below the horizontal line for a way to make this precise).
For a counterexample, consider a language with $\aleph_1$-many constant symbols $\{c_\alpha\mid \alpha<\aleph_1\}$ and the theory which says there are at least two elements in the domain and all the constants are equal: $T = \{\exists x\exists y\, x\neq y\}\cup \{c_\alpha = c_{\beta}\mid \alpha,\beta<\aleph_1\}$. Then $T$ is $\aleph_0$-stable, but $T$ is not axiomatizable by a countable set of sentences. If $|T'| = \aleph_0$, then $T'$ only mentions countably many constant symbols. Picking two constant symbols $c_\alpha$ and $c_\beta$ which are not mentioned in $T'$, we can find a model of $T'$ in which $c_\alpha\neq c_\beta$ by taking an arbitrary model of $T'$ and changing it to interpret $c_\alpha$ and $c_\beta$ to be two distinct elements of the domain. So  $T'\not\models T$.
This issue crops up in the first paragraph of your proposed proof: You say that if $c$ and $d$ are constants such that $T\models c = d$, then letting $T'$ be the theory obtained by replacing $d$ by $c$ everywhere, we have $T'\restriction_{\mathcal{L}\setminus \{d\}}\models T$. But that's not true, because in a model of $T'\restriction_{\mathcal{L}\setminus \{d\}}$, we can interpret the constant $d$ as any element, not necessarily $c$, so  $T'\restriction_{\mathcal{L}\setminus \{c\}}\not\models c = d$.
Also, as you note in the edit, what you've really proved in the fourth and fifth paragraphs of the proof is that there is a finite subset $\Delta(v)\subseteq  \Sigma(v)$ such that  $T\cup \Delta(v)\models \psi(v)$. What happens in the type space $S_1(T)$ is all happening modulo $T$, so the idea of focusing on the type space in one variable and then thinking of sentences in $T$ as formulas in one free variable isn't going to get you anywhere - any sentence in $T$ will automatically be in every type in $S_1(T)$, otherwise the type wouldn't be consistent with $T$! But actually, your idea of separating types and seeing that a small set of formulas suffice to define arbitrary formulas is on the right track... once you clarify what it is you should be trying to prove.

The correct statement is this: If an $L$-theory $T$ is $\kappa$-stable (in fact, $|S_n(T)|\leq \kappa$ for all $n$ is enough), then $T$ is a definitional expansion of an $L'$-theory $T'$ with $|L'|\leq \kappa$ (and this implies $|T'|\leq \kappa$). We can't hope to have $T'\models T$ in general, but still $T'$ is essentially the same as $T$.
Lemma: Suppose $\Delta$ is a set of $L$-formulas in $n$ free variables such that for any types $p\neq q$ in $S_n(T)$, there exists $\varphi\in \Delta$ such that $\varphi\in p$ and $\lnot \varphi\in q$. Then every $L$-formula in $n$ free variables is equivalent modulo $T$ to a finite Boolean combination of formulas in $\Delta$.
You can prove the lemma with two applications of the compactness theorem, along the lines of your attempted proof. But, for fun, here's a fancier topological proof. The formulas in $\Delta$ generate a Hausdorff topology on $S_n(T)$ which is coarser than the usual topology. If a Hausdorff topology on a set $X$ is coarser than a compact topology on $X$, then these topologies are equal. So $\Delta$ in fact generates the standard topology on $S_n(T)$. But in a Stone space (compact Hausdorff and totally disconnected) with a fixed clopen basis, every clopen set is a finite Boolean combination of the elements of the basis.
Ok, now let's fix $n$. For each pair $p\neq q$ in $S_n(T)$, let $\varphi_{p,q}$ be a formula distinguishing these types, and let $\Delta_n = \{\varphi_{p,q}\mid p\neq q\}$. By the Lemma, every formula in $n$ free variables is equivalent modulo $T$ to a finite Boolean combination of the formulas in $\Delta_n$.
Let $\Delta = \bigcup_{n\in \omega} \Delta_n$, and let $L'$ consist of all symbols in $L$ which appear in some formula in $\Delta$. Since $|S_n(T)| \leq \kappa$,  $|\Delta_n| \leq \kappa$, for all $n$, so $|\Delta|\leq \kappa$, and hence $|L'|\leq \kappa$.
For each $n$-ary relation symbol $R$ in $L\setminus L'$,  the formula $R(x_1,\dots,x_n)$ is equivalent modulo $T$ to a finite Boolean combination of formulas in $\Delta_n$, so it is definable by an $L'$-formula $\psi_R(x_1,\dots,x_n)$. Similarly, for every $n$-ary function symbol $f$, $f(x_1,\dots,x_n) = x_{n+1}$ is definable by an $L'$-formula $\psi_f(x_1,\dots,x_{n+1})$, and for every constant symbol $c$, $c = x_1$ is definable by an $L'$-formula $\psi_c(x_1)$.
Let $T'$ by the $L'$-formula obtained by replacing all symbols in $L\setminus L'$ by their definitions in the sentences of $T$. Then $T$ is a definitional expansion of $T'$, i.e. if we extend $T'$ by the definitions of all the symbols in $L\setminus L'$, the resulting theory is equivalent to $T$.
