If the Morley rank $RM(\phi)$ of a definable set $X$=$\phi(\mathcal{M})$ is $\alpha$, based on the inductive definition of RM (I am using David Marker's book Model Theory: An Introduction, definition 6.2.1. See the image below), we could form a sequence of successively strictly larger subsets of $X$ indexed by $\alpha$, wouldn't that mean the cardinality of $X$ is at least $\alpha$?
Also if the language size is $\lambda$, then the cardinality of all definable subsets is also $\lambda$, but the number of distinct definable subsets appears in the infinitely branching tree generated from the inductive definition of RM is at least $\alpha$.
There are examples such as DLO with countable model ($\mathbb{Q}$) and $\infty$ Morley Rank. What exactly am I missing here? It is not obvious to me where I am making a mistake.
The issue is that Morley rank (and all similar ranks) involve "counting down" rather than "counting up." And no matter how big an ordinal I start with, counting down in the ordinals will only last finitely many steps.
For example, taking $\mathcal{M}=(\mathbb{R};<)$ you can show with that every formula defining a nontrivial interval has Morley rank $\ge\alpha$ for every ordinal $\alpha$. Since intervals are definable from their (closure's) endpoints, each interval is definable by an $\mathcal{L}_M$-formula and so we can talk about intervals themselves in place of formulas. The Morley rank calculation here corresponds, not to any length-$\alpha$ sequence of sets for arbitrarily large $\alpha$, but rather to the fact that we can always partition a nontrivial interval into infinitely many nontrivial intervals.
(The same idea works for $(\mathbb{Q};<)$, we just need to replace "interval" with "interval whose closure's endpoints are rational" to deal with things like $(e,\pi)\cap\mathbb{Q}$.)
It may help to think of playing a game where at each round player $1$ supplies a partition of the interval under consideration and player $2$ picks one of the intervals in the partition and "winds down" an ordinal clock; player $1$ can keep this up for longer than player $2$, no matter what value the ordinal clock started at.