The answer for the second question is no. Truss proved in [1] that if we repeat Solovay's construction from a limit cardinal $\kappa$, we obtain a model in which the following properties:
- Countable unions of countable sets of real numbers are countable;
- Every well-orderable set of real numbers is countable;
- Every uncountable set of reals has a perfect subset;
- DC holds iff $\omega_1$ is regular iff $\kappa$ is inaccessible in the ground model;
- Every set of real numbers is Borel.
This shows that it is possible to have $CH+\aleph_1\nleq2^{\aleph_0}+\operatorname{cf}(\omega_1)=\omega$. However it does not answer the original (first) question.
Bibliography:
- Truss, John, Models of set theory containing many perfect sets. Ann. Math. Logic 7 (1974), 197–219.
While Gabe's answer is deleted ("One shouldn't try to work in ZF at 5am"), let me work in ZFC.
(a) The usual middle-thirds Cantor set $C$ is nowhere dense in $\mathbb R$. It has cardinal $\mathfrak c = 2^{\aleph_0}$. So every subset of $C$ is again nowhere dense in $\mathbb R$. Every subset of $C$ has the property of Baire. There are at least $2^{\mathfrak c}$ sets with the property of Baire. [In fact, there are exactly $2^{\mathfrak c}$ sets with the property of Baire.]
(b) Start with any family $\mathscr U$ of sets, and define
$$
\mathcal A(\mathscr U) = \{ \mathcal A(\mathcal X) :
\mathcal{X} = \langle X_s : s \in {}^{<\omega}\omega\rangle , X_s \in \mathscr U \text{ for all } s \in {}^{<\omega}\omega\} .
$$
The Suslin operation is idempotent. That is, if $\mathscr U$
is any family of sets, then $\mathcal A(\mathcal A(\mathscr U)) = \mathcal A(\mathscr U)$.
The family of "Suslin measurable sets" (a.k.a. coanalytic sets) is
$\mathcal A(\mathscr G)$, where $\mathcal G$ is the family of all open subsets of $\mathbb R$.
(c) Let $\mathscr G_0 = \{(a,b) : a,b\in\mathbb Q, a<b\}$. Then
$\mathcal A(\mathscr G) = \mathcal A(\mathscr G_0)$. Since $\mathscr G_0$ is countable and ${}^{<\omega}\omega$ is countable, there are at most $\aleph_0 ^{\aleph_0} = \mathfrak c$ Souslin schemes in $\mathscr G_0$. We conclude that $\mathcal A(\mathscr G_0)$ has cardinal at most $\mathfrak c$. So there are at most $\mathfrak c$ Suslin measurable sets in $\mathbb R$. [In fact, there are exactly $\mathfrak c$ Suslin measurable sets.]
(d) Conclude there is a set with the property of Baire that is not Suslin measurable.
Best Answer
In fact, the principle "Every set is analytic" is not consistent with $\mathsf{ZF}$ in the first place. We don't need choice to get a surjection $h$ from Baire space to the set of continuous maps on Baire space. But once we have such an $h$, the "diagonalizing" set $\{x: x\not\in h(x)\}$ can't be analytic.
Let me show how to get such an $h$ in $\mathsf{ZF}$ alone. Working in $\mathsf{ZF}$, let $C$ be the set of continuous maps $\omega^\omega\rightarrow\omega^\omega$. To each $\gamma\in C$ we assign the set $$S(\gamma)=\{(\sigma,\tau)\in\omega^{<\omega}\times\omega^{<\omega}: \forall f\succ \sigma(\gamma(f)\succ\tau)\}.$$ By your favorite coding mechanism we can identify $S(\gamma)$ with some $\hat{S}(\gamma)\in\omega^\omega$. But $\hat{S}(\gamma)=\hat{S}(\gamma')$ implies $\gamma=\gamma'$ for continuous $\gamma,\gamma'$, so in fact $\hat{S}$ gives an injection from $C$ to $\omega^\omega$.
Now turning a surjection into an injection without choice is hard, but the converse is trivial: for $x\in\omega^\omega$, let $h(r)=\hat{S}^{-1}(r)$ if $r\in ran(\hat{S})$, and let $h(r)$ be the always-zero map otherwise.