Let your manifold be $X=G/H$. First of all, since it is simply connected, we can write it as $K/U$ where $K$ and $U=K\cap H$ are compact in $G$ (Montgomery’s theorem, [1950](//ams.org/mathscinet-getitem?mr=37311)). Next, since $K/U$ is homogeneous symplectic, one knows that $U$ is the centralizer of a torus $S\subset K$.<sup>1)</sup> In particular $U$ contains any maximal torus containing $S$, i.e. $U$ is an equal rank subgroup of $K$. And finally, one knows that equal rank subgroups satisfy $χ(K/U)\ne0$: e.g. Samelson ([1958](//ams.org/mathscinet-getitem?mr=103509)), or Mostow ([2005](//ams.org/mathscinet-getitem?mr=2174096)).

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<sup>1) That is clear, with $S$ the closure of $\exp(\mathbf Rx)$, if we already know that $X\simeq$ the (co)adjoint orbit of some $x\in\mathfrak k^*\simeq\mathfrak k$. But it can also be proved *a priori* : Borel–Weil ([1954](http://www.numdam.org/item?id=SB_1951-1954__2__447_0), Thm 1).</sup>