set theory - Two questions about the boolean algebra $P(kappa)/Cub^*$

I have two questions based on exercises in Kunen's set theory. Let $kappa = cf(lambda) > omega$. Why is there a c.u.b. $C subseteq lambda$ of order type $kappa$? I thought we just choose $C$ as the image of an increasing unbounded function $kappa to lambda$, but I doubt that this has to be closed.

Also, why can we use this $C$ to get an isomorphism of boolean algebras $P(kappa)/Cub^*(kappa) cong P(lambda)/Cub^*(lambda)$? If we just pull back with $kappa to lambda$, I don't see why this will be well-defined. Note that $Cub^*$ is the ideal of non-stationary subsets.

The second problem is the following: Let $kappa$ be an uncountable regular cardinal. I want to prove that there is a decreasing sequence of stationary sets $S_alpha, alpha < kappa$, whose diagonal intersection is ${0}$. This is an exercise in Kunen's set theory, and there is a hint that one should use the preceding exercise, which says that the boolean algebra $B=P(kappa)/Cub^*(kappa)$ has infima indexed over $kappa$, which correspond to the diagonal intersection in $P(kappa)$.

Here's what I've done so far: Construct a decreasing sequence in $B$: Let $x_0=1$. If $x_alpha$ is already defined, define $x_{alpha+1} = x_alpha$ if $x_alpha$ is minimal and otherwise choose some $x_{alpha+1} < x_alpha$. If $alpha$ is a limit and $x_gamma$ is defined for all $gamma < alpha$, let $x_alpha$ be the infimum of these $x_gamma$.

Now if $x_alpha = [S_alpha]$, then $S_alpha$ is stationary iff $x_alpha neq 0$; is this the case? For $alpha < beta < kappa$, we have $x_beta leq x_alpha$, i.e. there is a c.u.b. $C_{alpha,beta}$ such that $S_beta cap C_{alpha,beta} subseteq S_alpha$. Now perhaps there is some double-index diagonal intersection $C$ of these $C_{alpha,beta}$ (?) which is c.u.b. again and which we may intersect with every $S_alpha$, so that we may assume $S_beta subseteq S_alpha$, as desired.

Finally we have to ensure the infimum of the $x_alpha$ is $0$. I wonder if this is true at all with this naive construction.