Let $\{S_n \mid n < \omega\}$ be a partition of $\aleph_1$ into countably many disjoint stationary subsets. Why adding a club of $\aleph_1$ to each $\aleph_1 \setminus S_n$ collapses $\aleph_1$ to $\aleph_0$?
Would it collapse $\aleph_1$ also adding single element to $\aleph_1 \setminus S_n$ for each $n < \omega$?
Or adding unbounded set for each $n$, not necessarily closed?
HINT: You add countably many clubs, suppose you didn't collapse $\aleph_1$, the intersection of all these clubs is still a club $A$. What do you know about $A\cap S_n$ and about the countable union $\bigcup(A\cap S_n)$?
For your second question, the answer is that you can do that without collapsing $\aleph_1$. Namely, given $S\subseteq\omega_1$ such that $\omega_1\setminus S$ is stationary, you can add a club which is disjoint from $S$ without collapsing $\aleph_1$. You can do that simply by looking at countable closed subsets which are disjoint of $S$, ordered by end-extension.
For the third question, you can just add any new subset of $\omega_1$, even one induced by adding a real, and look at how it translates to subsets of each $\omega_1\setminus S_n$.