If all finite subsets of a set have at most rank W, does the set itself necessarily have at most rank W?

Question

In other words, if no finite subset of a set has a rank greater than some ordinal, does the set itself have the same rank upper bound?

My proof sketch for yes is:

Assume the set S is non-empty and has no greatest rank element(trivial otherwise). Then the rank of S is a limit ordinal X which is the least ordinal greater than all members of S.

If the rank of any finite subset of S is at most V < X then since W is a limit ordinal there is an ordinal Y between V and X.

Either there is a subset of S with rank Y (contradiction), or Y is greater in rank than all members of S (also contradiction)

Does this work?

Answer 1

Either there is a subset of S with rank Y (contradiction), or Y is greater in rank than all members of S (also contradiction)

It looks to me like there is a step missing here (also, I suppose you mean "finite subset"?).

You can argue as follows. Let the rank of any finite subset of $S$ be not greater than some ordinal $\alpha$. Then in particular this holds for singleton subsets $\{x\}\subseteq S$. Now using the recursive definition of rank,

$$ rank(S)=\sup\{rank(x)+1\mid x\in S\}=\sup\{rank(\{x\})\mid x\in S\}\le\alpha. $$