[UPDATE] The question has been updated and is mostly unrelated to the question above the line. The updated question is below the line.
The following argument supports a "yes" answer; is it convincing?
Let $\Omega$ be the least uncountable ordinal. A finite ($L_{\omega \omega}$) formula of signature $\{\in\}$ of one free variable $\varphi(x)$ defines an element of $\Omega$ if $$(\Omega,\in) \models \exists!_x \varphi(x).$$ Let $X$ be the set of all $\varphi(x)$ such that $\varphi$ defines an element of $\Omega$. Since $L_{\omega \omega}$ is countable, so is $X$. Let $F\colon X\rightarrow \Omega$ be such that $F(\varphi)$ is the element of $\Omega$ such that $\varphi(F(\varphi))$. Let $F[X]$ be the range of $F$. Since $F[X]$ is a countable subset of $\Omega$, and since no countable subset of $\Omega$ is cofinal in $\Omega$, there exists a least $\beta\in\Omega$ such that $F[X]\subseteq\beta$. Since the preceding remarks can be formalized in ZFC and define $\beta$ uniquely, there exists a formula $\psi(x)$ such that $$(\Omega,\in)\models\forall_x \psi(x) \iff x = \beta.$$ Since $\psi\in X$ and $F(\psi)=\beta$, we have $\beta\in F[X]\subseteq \beta$, hence $\beta\in\beta$.
Question: is ZFC inconsistent?
As above, the following argument supports a "yes" answer.
Let $\Omega$ be the least uncountable ordinal. A finite $(L_{\omega \omega})$ sentence $\varphi$ of signature $\{\in\}$ selects an infinite countable ordinal if there exists an infinite countable ordinal $\alpha$ and set $x$ of rank $\alpha$ such that $$(\alpha\cup x,\in)\models\varphi.$$ Let $C$ be the set of all $\varphi$ such that $\varphi$ selects an infinite countable ordinal. Since $L_{\omega \omega}$ is countable, so is $C$. Let $Q\colon C\rightarrow\Omega$ be such that $Q(\varphi)$ is the least infinite countable ordinal $\alpha$ such that there exists a set $x$ of rank $\alpha$ such that $(\alpha\cup x,\in)\models\varphi$. Let $Q[C]$ be the range of $Q$. Since $Q[C]$ is a countable subset of $\Omega$, and since no countable subset of $\Omega$ is cofinal in $\Omega$, there exists a least $\beta\in\Omega$ such that $Q[C]\subseteq\beta$. Is there a $\psi\in C$ such that $Q(\psi)=\beta$?
Case 1: yes. We have $Q(\psi)\in Q[C]\subseteq\beta$, hence $\beta\in\beta$, a contradiction.
Case 2: no. In other words, for all $\varphi\in C$, $Q(\varphi)\not=\beta$. If there exists a $\varphi\in C$ such that $Q(\varphi)>\beta$, then we would have $\beta\in Q(\varphi)\in Q[C]\subseteq\beta$, a contradiction. Therefore, for all $\varphi\in C$, $Q(\varphi)<\beta$. Let $\psi$ be the sentence, "there exists an infinite countable ordinal $\gamma$ and set $u$ such that for all $\varphi\in L_{\omega \omega}$, if $\varphi$ selects an infinite countable ordinal, then $(\gamma\cup u,\in)\models\varphi$. Claim: $Q(\psi)=\beta$, a contradiction.
[UPDATE] Replace the definition of $\psi$ with "there exists an infinite countable ordinal $\gamma$ such that for all $\varphi\in L_{\omega \omega}$, if $\varphi$ selects an infinite countable ordinal, then there exists a set $u$ such that $(\gamma\cup u,\in)\models\varphi$."