$\kappa$-closed forcing preserves stationary sets. Let's take an uncountable cardinal $\kappa$ which is regular inside the ground model $M$ and $\mathbb{P}\in M$ a forcing notion which is $\kappa-$closed in $M$. I'm trying to prove that every stationary set $S$ in $M$ continues to be stationary in the generic extension $M[G]$. So, suppose $S$ wouldn't be a stationary set in $M[G]$, then there would exists a condition $p_0\in G$ and a club set $C$ (in $M[G]$) such that $p\Vdash \check{S}\cap \dot{C}=\emptyset\;\wedge\;\dot{C}\;\text{club}$
I've applied that argument using $\kappa-$closeness by which it's possible to construct a sequence $\{\beta_\alpha\}_{\alpha<\kappa}$ of ordinals and $\{p_\alpha\}_{\alpha<\kappa}$ such that $p_\alpha \Vdash\check{\beta_\alpha}\in \dot{C}$ and $p_0\leq p_\alpha$ for every $\alpha\in\kappa$. 
But from this point I can't follow the argument. I mean, why there exists some $\beta_\alpha\in S$? Is it due to stationariety of $S$ in $M$? If it would be the case, why $\{\beta_\alpha\}_{\alpha<\kappa}$ is club in $M$?
If this questions would answered then $\mathbb{1}\Vdash \check{\beta_\alpha}\in \check{S}$ and then $p_\alpha\Vdash \check{S}\cap \dot{C}\neq\emptyset$ which would give us the desired contradiction.
 A: In the following, I use the convention that $p \le q$ means "$p$ is stronger than $q$".
For every $p \in \Bbb P$ such that $p \not \Vdash \dot {C} \text{ is not a club in } \check{\kappa}$, we may recursively construct a sequence
$( (p_{\alpha}, \beta_{\alpha}) \mid \alpha < \kappa$ of
$p_{\alpha} \in \Bbb P$ and $\beta_{\alpha} < \kappa$ as follows: Let
$p_{0} \le p$ and $\beta_{0} < \kappa$ be such that
$p_{0} \Vdash \dot C \subseteq \check{\kappa} \text{ is club and }
\check{\beta}_{0} = \min \dot{C}$. Given $(p_{\alpha}, \beta_{\alpha})$
fix $p_{\alpha +1} \le p_{\alpha}$ and
$\beta_{\alpha} < \beta_{\alpha+1}$ such that
$p_{\alpha+1} \vDash \check{\beta}_{\alpha+1} \in \dot{C}$. Now let
$\lambda < \kappa$ be a limit ordinal such that
$((p_{\alpha}, \beta_{\alpha}) \mid \alpha < \lambda )$ has been
constructed. Fix $p_{\lambda}$ such that $p_{\lambda} \le p_{\alpha}$
for all $\alpha < \kappa$ and let
$\beta_{\lambda} := \sup \{ \beta_{\alpha} \mid \alpha < \lambda
\}$. Since $p_{\lambda} \Vdash \check{\beta}_{\alpha} \in \dot{C}$ for
all $\alpha < \lambda$ and
$p_{\lambda} \Vdash \dot{C} \text{ is club}$, we have that
$p_{\lambda} \Vdash \check{\beta}_{\lambda} \in \dot C$. Now
$(\beta_{\alpha} \mid \alpha < \kappa)$ is strictly increasing,
continuous and cofinal in $\kappa$ and hence
$\{ \beta_{\alpha} \mid \alpha < \kappa \}$ is club (in $M$). Thus
there is some $\alpha < \kappa$ such that $\beta_{\alpha} \in S$. In
fact, $p_{\alpha} \Vdash \dot C \cap \check{S} \neq \emptyset$.
There is hence no $p \in \Bbb P$ such that $p \Vdash \dot C \text{ is
  club in } \check{\kappa} \text{ and } \check{S} \cap \dot{C} = \emptyset$ and therefore $S$ remains to be stationary in $M[G]$.
