I don't know if the proof of the theorem below is correct. The goal is to do the proof from the beginning, without using compactness or Löwenheim-Skolem. Only the Tarski-Vaught test (maybe).
We are working in first-order logic with a language $\mathcal{L}$, so the set of all formulas, $\text{Form}(\mathcal{L})$, has cardinality $|\mathcal{ L}|+|\text{Var}|$, where $\text{Var}$ is the set of variables, which we will take countable, so $|\text{Var}|=\aleph_0$. Furthermore, we say that the formula $\varphi$ is satisfiable if $\mathcal{M}\models\varphi[a_1,\dots,a_n] $ for some $a_1,\dots,a_n\in M$, where the notation $ [a_1,\dots,a_n]$ indicates the substitution of the free variables of $\varphi$ by the elements $a_1,\dots,a_n$ (which we will call parameters of $\varphi$) of the universe of the structure $\mathcal{M}$. We say that a set of formulas has a model if all its formulas are satisfiable for the same structure $\mathcal{M}$ and the corresponding substitution.
Theorem. If $\Sigma\subset\text{Form}(\mathcal{L})$ has a model, then it has a model of cardinality $|\mathcal{L}|+\aleph_0$.
Proof. Consider some infinite model $\mathcal{M}$ of $\Sigma$, we want to find some substructure $\mathcal{N}$ of $\mathcal{M}$ that is also a model of $\Sigma$. For every formula $\varphi\in\Sigma$, we build the set $M_\varphi$ as follows: if $\varphi$ has free variables $x_1,\dots,x_n$, we pick a set of parameters $\{a_1,\dots,a_n\} $ such that $\mathcal{M}\models\varphi[a_1,\dots,a_n]$, if $\varphi$ is of the form $\varphi=\exists x_1\dots\exists x_m\psi$, we take a set of witnesses $b_1,\dots,b_m$; then, $M_\varphi=\{a_1,\dots,a_n,b_1,\dots,b_m\}$. Note that, if $\varphi$ is a sentence and quantifier free, $M_\varphi=\varnothing$ (later, we will add constants and function symbols to take this case into account). Now, we consider the set $ A=\bigcup_{\varphi\in\Sigma}M_\varphi $. Since $|\Sigma|\leq|\mathcal{L}|+\aleph_0$ and $|M_\varphi|<\aleph_0$ for every $\varphi\in\Sigma $, we have $|A|\leq|\mathcal{L}|+\aleph_0$.
Now, we take the union $B=A\cup \{c^\mathcal{M}\in M:c \text{ is a constant of }\mathcal{L}\}$. Let's define $B_0=B$ and $$B_{n+1}=B_n\cup\bigcup_{f\in\mathcal{F}}\text{Im}(f^\mathcal{M}\upharpoonright B_n^{n_f}),$$ where $\mathcal{F}$ is the set of function symbols of $\mathcal{L}$, $n_f$ is the arity of $f$ and $\text{Im}(f^\mathcal{M}\upharpoonright B_n^{n_f})$ is the image of the restriction of $f^\mathcal{M}$ to $B_n^{n_f}$. Note that $|B_0|\leq|\mathcal{L}|+|\mathcal{L}|+\aleph_0=|\mathcal{L}|+\aleph_0$ and $|\bigcup_{f\in\mathcal{F}}\text{Im}(f^\mathcal{M}\upharpoonright B_n^{n_f})|\leq|\mathcal{F}|\cdot|B_n|\leq|\mathcal{L}|\cdot|B_n| $, which proves, by induction, that $|B_{n+1}|\leq |\mathcal{L}|+\aleph_0$. Therefore, $C=\bigcup_{n\in\omega} B_n$ has cardinality $|\mathcal{L}|+\aleph_0$ at most. Moreover, by construction, it is the universe of a substructure of $\mathcal{M}$ which is a model of $\Sigma$, so we are done.
Is there any mistake in that proof?
What if there is no infinite model $M$ of $Σ$, Only finites? For example let $Σ=\{∀x\forall y (x=y)\}$ has only models of cardinality $1$, certainly it has no models of cardinality $|{\cal L}|+\aleph_0=\aleph_0>1$.
The correct theorem needs the added assumption that $Σ$ has a model of cardinality $≥|{\cal L}|+\aleph_0$.
The rest of your proof has the right idea, but it doesn't quite work.
First of all, your $M_φ$ doesn't take into account "dependencies" in existential quantifiers (look e.g. on $φ(x)=∃y (y<x)$, your $M_φ$ will be a pair $\{a,b\}$ such that $b<a$, but then $φ(b)$ doesn't hold and requires a different witness) because of this "lack of dependency" your resulting $C$ need not satisfy $Σ$, and even if you fix that, after going from $A$ to $B_0$, you add even more unrelated elements, so you again "break" some sentences.
One can fix the definition of $M_φ$ by making it a function instead of a constant:
Given a formula $φ(\overline x,y)$ and $\overline c∈M^n$ such that $M\models ∃x φ(\overline c, x)$, define $f_φ(\overline c)$ to be some witness of this (for $\overline c$ such that $M\not\models ∃x φ(\overline c, x)$, $f_φ$ is not defined). This $f_φ$ is called a Skolem function of $φ$ (note that the Skolem function indeed depends on $M$, but it is omitted for brevity, and also note that for a given $φ$ we may have a lot of different Skolem functions of $φ$, but we just choose one such function). For brevity, if $φ$ has no free variables apart from $x$ and $M\models ∃x φ(x)$, we consider $f_φ$ to be a "0-arity" function.
Now let $F=\{f_ψ\mid M\models ∃\overline y ∃x φ(\overline y,x)\}$ be a set of Skolem functions. Those Skolem functions have the same idea as your $M_φ$, but the key difference is that it adds the dependency of the parameter.
($f''x$ is the image of $f$ under $x$. In the follows ignore arity and domain problems, whenever $f$ is a function of arity $n$ with domain $X$, when I write $f''Y$ I actually means $f'' (X∩Y)^n$. We additionally say that if $f$ is a "0-arity" function as defined above, the image of $f$ under any set is just the singleton it represent)
So we start our construction with $B_0=∅$. Now we want to make $B_0$ satisfy all of the formulaes of the form $∃x φ(\overline c,x)$ for $\overline c∈B_0$, so we define $B_1=\bigcup_{f\in F} f''B_0$ (remember, from our abuse of notation, this means that $B_1$ is the set of constants representative we chose for formulaes with a single free variable).
But now we want to make $B_1$ to also satisfy all of those formulaes, so we define $B_2=\bigcup_{f\in F} f''B_1$, and so on and so forth, till we define $B_ω=\bigcup_{n\in\mathbb N} B_n$.
This $B_ω$ is called "the Skolem Hull" of $M$ (note that the term here is a bit confusing, we say it is the Skolem Hull, but this Hull depends on our choice of Skolem functions).
We can see that $|B_0|=0,|B_n|≤|Form({\cal L})|+|B_{n-1}|$, so $|B_ω|≤|Form({\cal L})|$ as it is countable union of $≤|Form({\cal L})|$ sets and $|Form({\cal L})|$ is infinite.
Now you need to show that $B_ω$ satisfy $Σ$, to do this you prove the stronger property of: $B_ω$ is an elementary substructure of $M$, to see this you can use Tarski-Vaught criterion and see that it is immediate from the definition of our Skolem functions