 Solution

An inner model $M$ is projectively well-ordered when some projective relation well-orders the real numbers of $M$. The ordinal ${\aleph }_1^M={\omega }_1^M$ is the least ordinal that $M$ regards as uncountable; equivalently, it is the supremum of the order types of the well-order codes in $M$.

Assume for contradiction that ${\omega }_1^M$ is uncountable in the ambient universe. Use the projective well-order of the reals of $M$ to choose, for each $\alpha <{\omega }_1^M$, the least $M$-real coding a well-order of type $\alpha$. Standard closure properties of the projective hierarchy make the resulting set $U\subseteq \mathrm{WF}$ projective. It is uncountable because it contains one distinct code for every $\alpha <{\omega }_1^M$.

The set $U$ has no perfect subset. Indeed, a perfect subset $P\subseteq U$ is closed and therefore analytic. The boundedness theorem for well-order codes bounds the ranks of its members below one countable ordinal $\beta$. Since $U$ contains at most one code of each rank, $P$ would then be countable, whereas every nonempty perfect set of reals is uncountable.

If every projective set is determined, projective determinacy holds and gives the perfect set property to every projective set. Applying it to the uncountable projective set $U$ yields a perfect subset, a contradiction. ThereforeThis is projective determinacy collapses the inner-model omega-one.