(Contradiction) All symplectic manifolds are holomorphic I’m studying symplectic manifolds and almost complex structures. This lead to two propositions:
Proposition 1 (from da Silva’s Lectures on Symplectic Geometry): If $J_0$ and $J_1$ are almost complex structures compatible with a symplectic manifold $(M,\omega)$, then there is a family of almost complex structures $J_t, t \in [0,1]$ such that $J_t$ is compatible with $(M,\omega)$.
Proposition 2 (from Hatcher’s K-theory book): Let $E \twoheadrightarrow B \times [0,1]$ be a (topological) vector bundle. Then the restrictions of $E$ to $B \times \{0\}$ and $B \times \{1\}$ are isomorphic.
I think proposition 2 can be extended to smooth bundles, though maybe this is the source of my confusion below.
The confusion: I think I can prove the following absurdity using the above two propositions.
Absurdity: Let $(M^{2n},\omega)$ be a symplectic manifold. Then $M$ admits a holomorphic atlas.
Proof: Let $\phi: U \to \mathbb{R}^{2n}$ be a Darboux chart (the pullback of the standard symplectic form on $\phi(U)$ is $\omega$) of a neighborhood $U \subset M$ around an arbitrary point $p \in U$. Let $J$ be an almost complex structure on $TM$. We have two almost complex structures on $\phi(U)$. The first one, call it $j_0$, comes from $J$:
$$ j_0 \equiv \phi_* \circ j \circ \phi^{-1}_*.$$
The second one, call it $j_1$, comes from identifying $\mathbb{R}^{2n}$ with $\mathbb{C}^n$. Since $\phi$ is a Darboux chart, $j_0$ is compatible with the standard symplectic form. $j_1$ is also compatible with the standard form. Thus, Proposition 1 tells us that we have a family $j_t$ of almost complex structures on $\phi(U)$ compatible with the standard symplectic form.
We now construct a complex vector bundle $\pi: E \twoheadrightarrow \phi(U)$ as follows. For a point $(x,t) \in \phi(U) \times [0,1]$, the fiber $\pi^{-1}(x,t)$ is the vector space $T_x \phi(U)$ equipped with the almost complex structure $j_t$. From Proposition 2, we obtain an isomorphism $\psi: \left(\phi(U),j_0\right) \to \left(\phi(U),j_1\right)$. Notice $(\phi(U),j_1)$ is a neighborhood in $\mathbb{C}^n$. Therefore, $\psi \circ \phi$ is a holomorphic chart around $p$.
$p$ is arbitrary, so we obtain our atlas. $\blacksquare$
I’ve been perusing this proof for a while, but I cannot clearly see where it fails. What am I missing?
 A: The comments of @JHM and @Ivan Solonenko contain an answer to OP's question.
(1) The OP is correct that $j_0:=\phi_* \circ J \circ \phi^{-1}_*$ is an almost-complex structure on the image $V:=\phi(U) \subset \mathbb{R}^{2n}$ which is compatible with the standard symplectic form $\omega_{std}$ satisfying $\omega=\phi^* \omega_{std}$, where $(U, \omega, J)$ is the initial pseudo-holomorphic structure.
(2) Since $\phi$ is a Darboux chart, then $\omega_{std}$ is also compatible with the standard almost complex structure $j_1$ on $\mathbb{R}^{2n}$ restricted to $V$.
So we have two $\omega_{std}$-compatible a.c. structures $j_0, j_1$ on $V$, and can even find a smooth $1$-parameter family of compatible a.c  structures $j_t$ on $V$, for $0\leq t \leq 1$.
(3) The OP constructs a complex vector bundle $E \to V \times [0,1]$, where the fibre over $(p,t)$ is the $j_t$-complex vector space over $T_{p} V$. Now the OP correctly applies a form of Hatcher's Prop.2, obtaining a complex vector bundle isomorphism $F: E|_{V \times \{0\}} \to E|_{V \times \{1\}}$.
(4) As clearly stated by @Ivan Solonenko, here the OP errs in interpreting $F$ as the differential of an isomorphism between the base space $V=\phi(U)$. Indeed $F$ is the identity map on the base $V$. The existence of fibrewise complex isomorphisms does not imply that the base spaces $(V, j_0)$ and $(V, j_1)$ are holomorphic. If the isomorphism $F$ produced by Prop.2. was induced by a diffeomorphism between the bases $f: V \to V$, then we would have $F=df$ on the fibres, and would find $df\circ j_0=j_1\circ df$. But this is not the case.
In conclusion, fibrewise deformations like $F$ are not induced by maps on the base. $F$ is not a differential $df$ except if integrability conditions are satisfied. For a.c. structures, the tensorial form of the integrability condition is given by Nijenhuis tensor. E.g., the $\pm i$-eigenspaces of a.c. structures $j|_p$ define distributions on $V$, and which are almost never integrable for generic a.c. structures.
This can be recurring point of frustration in studying symplectic geometry, and is basically the subject of Gromov-Eliashberg $h$-principles. You might find Eliashberg--Mishachev's AMS textbook "Introduction to $h$-principle" interesting -- it contains several such examples as above.
