[step:Trivialize the smooth family near the central fibre]
Regard $X:=X_0=\pi^{-1}(0)$ as the smooth manifold underlying the central fibre, and let $J_0:TX\to TX$ be its complex structure. Choose a Kähler form $\omega_0$ on $(X,J_0)$, so $\omega_0$ is a closed real smooth $2$-form of type $(1,1)$ and $\omega_0(v,J_0v)>0$ for every nonzero tangent vector $v\in TX$. Write $A^{1,1}(X,J_0;\mathbb R)$ for the real smooth $(1,1)$-forms on $(X,J_0)$ and $A^2(X;\mathbb R)$ for the real smooth $2$-forms on $X$.
By Ehresmann's proper submersion theorem, applied to the proper smooth submersion underlying $\pi$, there is an open neighbourhood $B_1\subset B$ of $0$ and a smooth diffeomorphism $\Phi:X\times B_1\to \pi^{-1}(B_1)$ satisfying $\pi(\Phi(x,b))=b$ for all $(x,b)\in X\times B_1$ and $\Phi(x,0)=x$ after identifying $X$ with $\mathcal X_0$. For each $b\in B_1$, define the smooth fibre-identification map $\Phi_b:X\to \mathcal X_b$ by $\Phi_b(x)=\Phi(x,b)$. Pulling the complex structure of $\mathcal X_b$ back by $\Phi_b$ gives a smooth family of integrable almost-complex structures $J_b:TX\to TX$, with $J_0$ equal to the original complex structure on $X$.
The pulled-back form $\omega_0$ is a fixed closed real $2$-form on the smooth manifold $X$, so in particular $\omega_0\in A^2(X;\mathbb R)$. To prove that $\mathcal X_b$ is Kähler, it is enough to construct, for all $b$ sufficiently close to $0$, a closed real $2$-form $\omega_b$ on $X$ that has type $(1,1)$ with respect to $J_b$ and is positive with respect to $J_b$, because then $(\Phi_b^{-1})^*\omega_b$ is a Kähler form on $\mathcal X_b$.
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