Prove in $\text{ZFC}$ that if there is some ordinal $\alpha$ with $V_\alpha \models \text{ZFC}$, then $\text{cf}(\beta)=\omega$ where $\beta$ equals the least such $\alpha$.
Here is what I came up with.
Suppose that $\text {cf}(\beta)>\omega$. Then $|V_{\beta}|> \aleph_0$.
Apply the Lowenheim-Skolem theorem (L-S) to get an elementary substructure $E_0 \preceq V_{\beta}$ where $|E_0|=|V_{\omega}|$. Then there is some least $b_0 < \beta$ such that $E_0\subset V_{b_0}$.
Apply L-S again to get $V_{b_0}\subset E_1 \preceq V_{\beta}$ where $|E_1|=|V_{b_0}|<|V_{\beta}|$. We can find some least $b_1<\beta$ such that $V_{b_1}\supset E_1$. Continuing like this, we get $E_0\subset V_{b_0}\subset E_1 \subset V_{b_1}\subset E_2 \subset \cdots $, with $E_i\preceq V_{\beta}$ for each $i\in \mathbb N$. Then $\bigcup_n E_n = \bigcup_n V_{b_n} = V_{\sup\{b_n\}}$. As $E_i \models \text{ZFC}$ for each $i$, it follows that $V_{\sup\{b_n\}} \models \text{ZFC}$.
The problem is that I'm not sure how to justify that at each point in the chain (after the first), we can find a $V_{b_i}$ that contains $E_i$. I think it's true but am not sure and can't give a rigorous argument. I would appreciate any help.
You're correct that this is not clear that you can find such $\beta_i$, since who's to say that $E_i$ is bounded below $\beta$?
But you can solve this by going a similar but slightly different route: Reflection.
Since $V_\beta$ is a model of $\sf ZFC$, the reflection theorem holds for it, and note that the von Neumann hierarchy of $V_\beta$ is made of actual $V_\alpha$'s and not just parts of them. Now by induction reflect the first $n$ axioms of $\sf ZFC$ on higher and higher ordinals. By the fact we do this in $V$, there is no confusion with meta-theoretical issues (namely, the meta-theory of $V_\beta$ is $V$ itself), and by uncountable cofinality we can find a limit point where all $\sf ZFC$ axioms are reflected.
Therefore if $\beta$ had uncountable cofinality, it wasn't the first.