[step:Use holomorphicity of the inclusion to preserve real type $(1,1)$]
Since $Y\subset X$ is a complex submanifold, the inclusion $\iota:Y\to X$ is holomorphic. Therefore, for every $y\in Y$, the differential $d\iota_y:T_yY\to T_{\iota(y)}X$ is complex-linear, meaning $d\iota_y\circ J_Y=J_X\circ d\iota_y$.
Because $\omega$ is a real $(1,1)$-form on $X$, it is real-valued and satisfies the $J_X$-invariance identity $\omega_{\iota(y)}(J_X a,J_X b)=\omega_{\iota(y)}(a,b)$ for all $y\in Y$ and all $a,b\in T_{\iota(y)}X$. Let $u,v\in T_yY$. By the definition of pullback and complex-linearity of $d\iota_y$, $\alpha_y(J_Yu,J_Yv)=\omega_{\iota(y)}(d\iota_y(J_Yu),d\iota_y(J_Yv))$. Using $d\iota_y(J_Yu)=J_Xd\iota_y(u)$ and $d\iota_y(J_Yv)=J_Xd\iota_y(v)$, this becomes $\alpha_y(J_Yu,J_Yv)=\omega_{\iota(y)}(J_Xd\iota_y(u),J_Xd\iota_y(v))$. The $(1,1)$ identity for $\omega$ gives $\alpha_y(J_Yu,J_Yv)=\omega_{\iota(y)}(d\iota_y(u),d\iota_y(v))=\alpha_y(u,v)$. Thus $\alpha$ has type $(1,1)$ on $Y$. Since $\omega$ is real-valued and pullback preserves real-valuedness, $\alpha$ is a real $(1,1)$-form.
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