Call a cardinal $\kappa$ worldly iff $V_\kappa\vDash ZFC$. Let $\kappa_\alpha$ be the $\alpha$th worldly cardinal, i.e. the least worldly cardinal such that $\{\beta\lt\kappa_\alpha|\beta\text{ is worldly}\}$ has order-type $\alpha$. Since the class of worldly cardinals are closed, then $\kappa_\omega=\sup\kappa_n$.
Therefore, $V_{\kappa_\omega}\vDash\forall\alpha(\exists\delta\gt\alpha(V_\delta\vDash ZFC))$. Now define $f: \omega\rightarrow\kappa$ by $f(n)=\kappa_n$. We have that $f(n)$ is definable and absoloute to $V_{\kappa_\omega}$. Therefore, $f"\omega\in V_{\kappa_\omega}$ by the axiom of replacement. But this is impossible.
Best Answer
The worldly cardinals are not closed, and you have proved this. You are probably remembering the similar-sounding fact that $ \{\alpha: V_\alpha\prec V_\kappa\}$ is a club in $\kappa$ for worldly $\kappa$ of uncountable cofinality. (And note that the argument for closure relies on an elementary chain argument.) Picking out this cofinal sequence requires $V_\kappa$ know what is elementary to it, which it doesn't. And nor does $V_\alpha$ for the $\omega$-th such $\alpha$ know which worldly cardinals beneath it are elementary submodels of $V_\kappa.$