The answer to Question 2 is yes; such a chain can be found using only diagonal operators with respect to some fixed basis $(e_n)_{n\in{\mathbb N}}$ for $H$. As a starting point, you can find an $\omega_1$-sequence of subsets $S_\alpha\subset{\mathbb N}$ such that if $\alpha<\beta$, $S_\beta\setminus S_\alpha$ is finite and $S_\alpha\setminus S_\beta$ is infinite (that is, the sequence $(S_\alpha)$ is strictly decreasing "modulo finite sets"). This is easy to do because by diagonalizing, given a countable decreasing sequence of infinite subsets of ${\mathbb N}$, you can always find another infinite set which is strictly contained in all of them modulo finite sets. Now let $A_\alpha$ denote the algebra of diagonal operators such that the eigenvalues of the eigenvectors $e_n$ for $n\in S_\alpha$ form a convergent sequence. These are closed because a uniform limit of convergent sequences is convergent. These are nested because if $\alpha<\beta$, then all but finitely many elements of $S_\beta$ are contained in $S_\alpha$ so if the $S_\alpha$-eigenvalues converge, so do the $S_\beta$-eigenvalues.
Here's a more conceptual explanation of this construction. The algebra of diagonal operators is naturally isomorphic to the algebra $C_b({\mathbb N})$ of bounded continuous functions on $\mathbb N$, which is in turn naturally isomorphic to the algebra $C(\beta{\mathbb N})$ of all continuous functions on the Stone-Cech compactification of $\mathbb N$. Each set $S_\alpha\subset{\mathbb N}$ has a closure $\overline{S_\alpha}\subset\beta{\mathbb N}$, and the statement that the $S_\alpha$ are decreasing modulo finite sets says exactly that the sequence of closed sets $C_\alpha=\overline{S_\alpha}\setminus S_\alpha$ is actually decreasing. The subalgebra $A_\alpha$ is then exactly the set of continuous functions on $\beta{\mathbb N}$ which are constant on the set $C_\alpha$.