How can we show that the intersection of a series of closed unbounded sets is closed unbounded? I am currently stuck on an exercise in set-theory and would like to get some help: If $S$ is a stationary subset of a regular uncountable cardinal $κ$, and a subset $C$ of $κ$ is an S-cub if it is unbounded in $κ$ and $\sup(x) ∈ C$ holds for every $x ⊆ C$ with $\sup(x) ∈ S$, then how can we prove that for every sequence $(C_α | α < λ)$ of S-cubs with λ < κ, $\bigcap_{α<λ} C_α$ is an S-cub? Thanks in advance.

It is clear that we satisfy the closure part, since if a bounded set $x$ with $\sup(x)\in S$ is contained in the intersection $\bigcap_\alpha C_\alpha$, then $x$ is contained in each $C_\alpha$, and so $\sup(x)$ is in each $C_\alpha$ and hence in the intersection $\bigcap_\alpha C_\alpha$, as desired.

So the only difficult part is to show that $\bigcap_\alpha C_\alpha$ is unbounded in $\kappa$. For this, let $\bar C_\alpha$ be the closure of $C_\alpha$, that is, $C_\alpha$ with all of its limit points. So this is a club set in $\kappa$, and since the intersection of fewer than $\kappa$ many clubs is club, it follows that $\bigcap_\alpha\bar C_\alpha$ is closed and unbounded in $\kappa$. Thus, since $S$ is stationary, there are unboundedly many $\beta\in S$ that are limits of $\bigcap_\alpha \bar C_\alpha$, and these $\beta$'s must all be in every $C_\alpha$ since each $C_\alpha$ is $S$-closed.