 Solution

The induced ordering on $x$ is a well-order. By transfinite recursion define $h(t)=\{h(s):s\in x,s<t\}$ for $t\in x$. Induction shows that each $h(t)$ is an ordinal, that these values strictly increase, and that their range is the ordinal $\mu (x)={\bigcup }_{t\in x}(h(t)+1)$. The map $h$ is an order isomorphism from $x$ to this ordinal. Any order isomorphism between ordinals must fix their elements successively, so this ordinal is unique.

An increasing map $j:\kappa \to \lambda$ between ordinals satisfies $j(\xi )\ge \xi$ by induction: its predecessors include $j(\eta )$ for all $\eta <\xi$. Hence $\kappa \le \lambda$. Apply this to the inclusion $x\subset y$, transported through their order isomorphisms, and then to $y\subset \alpha$. It follows that $\boxed{\mu (x)\le \mu (y)\le \alpha }$.

If each $x_n$ is an initial segment of later $x_m$, uniqueness makes their ordinal enumerations compatible: the enumeration of $x_m$ restricts to that of $x_n$. Their union is consequently an order isomorphism from ${\bigcup }_n{x}_n$ to ${\bigcup }_n\mu (x_n)$. This proves the asserted equality. Without the initial segment condition it fails: in $\alpha =\omega +1$, take $x_n=\{0,\dots ,n-1\}\cup \{\omega \}$. Then $\mu (x_n)=n+1$, whose union is $\omega$, but $\mu ({\bigcup }_n{x}_n)=\omega +1$.

For decreasing sets, $\mu (x_n)$ is a nonincreasing sequence of ordinals. Its set of values has a least member, attained at some $n_0$; all later values must equal it. Hence the sequence is eventually constant. The intersection formula nevertheless fails: take $x_n=\{n,n+1,\dots \}\subset \omega$. Every $x_n$ has order type $\omega$, whereas their intersection is empty and has order type zero. Thus $\mu ({\bigcap }_n{x}_n)=0\ne {\bigcap }_n\mu (x_n)=\omega$.