= 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 $x_n=\{0,\ldots,n-1\}\cup\{\omega\}$. Then $\mu(x_n)=n+1$, whose union is $\omega$, but $\mu(\bigcup_nx_n)=\omega+1$.
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 \b[eventually constant]. The intersection formula nevertheless fails: take $x_n=\{n,n+1,\ldots\}\subset\omega$. Every $x_n$ has order type $\omega$, whereas their intersection is empty and has order type zero. Thus $\mu(\bigcap_nx_n)=0\ne\bigcap_n\mu(x_n)=\omega$.
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