= Solution
The assumed <beth number> equality $\beth_1=\aleph_2$ means $2^{\aleph_0}=\aleph_2$ in the ambient universe. Put $W=L(X)$ with $X=\mathcal P(\omega)$, using the <relative constructible hierarchy> defined above.
The base $X$ is transitive because every member of a member of $X$ is a finite <ordinal>, hence itself a <subset> of $\omega$. Moreover $X\subseteq W$ and $X\in W$, and $W$ is an <inner model> of <ZF>. The latter standard fact follows from the hierarchy and reflection: bounded-rank witnesses give pairing and union; reflection gives separation and replacement; ambient replacement bounds the stages of all relatively constructible <subsets> of any fixed <set>, giving power set. It does not require <axiom of choice> inside $W$. Since no new <subsets> of $\omega$ can appear in an inner class,
$$
\mathcal P^W(\omega)=X=\mathcal P(\omega).
$$
Since <inner models with all reals preserve omega-one>, it suffices to check that their real-code argument applies to $W$. Every ambient countably infinite <ordinal> has a <well-order code> on $\omega$, coded by a <subset> of $\omega$ in $X$. Finite <ordinals> and $\omega$ are already shared. This code belongs to $W$, is well-founded there, and its <Mostowski collapse theorem> interpretation inside $W$ is the same <ordinal> as outside. Hence the <ordinal> is countable in $W$. Conversely, any countability witness in $W$ remains a witness in the ambient universe. Thus $\omega_1^W=\omega_1$.
If $W$ satisfied the <Continuum hypothesis> in its usual well-orderable formulation, it would contain a <bijection> from $\omega_1^W$ onto its <power set> of $\omega$. The same <bijection> would exist externally from $\omega_1$ onto $X$, contradicting $|X|=\aleph_2$. Therefore
$$
\boxed{L(\mathcal P(\omega))\models\neg\mathsf{CH}.}
$$
This proof uses agreement on all reals and on $\omega_1$, not an unwarranted assertion that $L(X)$ satisfies <axiom of choice> or computes every higher <cardinal number> correctly.
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