[proofplan]
Pull the Kähler form back to $Y$ and observe that the restriction remains a positive real $(1,1)$-form on the complex tangent spaces of $Y$. In local holomorphic coordinates on $Y$, the $m$-fold wedge of this restricted form is a positive smooth multiple of the complex orientation volume form. Since $Y$ is compact and nonempty, integrating a positive smooth top-degree form over $Y$ gives a strictly positive number. The cohomological statement follows because integration of a closed top-degree representative over $Y$ computes the pairing with the fundamental class.
[/proofplan]
[step:Restrict the Kähler form to the complex submanifold]
Let $\iota:Y\hookrightarrow X$ denote the inclusion map. Let $J_X:TX\to TX$ and $J_Y:TY\to TY$ denote the complex structures on $X$ and $Y$, respectively. Define
\begin{align*}
\eta := \iota^*\omega \in \Omega^2(Y)
\end{align*}
to be the pullback of $\omega$ to $Y$. Since $\omega$ is closed, $d\eta = d(\iota^*\omega) = \iota^*(d\omega) = 0$. Since $\iota$ is holomorphic and $Y$ is a complex submanifold, $\eta$ is a real $(1,1)$-form on $Y$.
For every point $p \in Y$ and every nonzero tangent vector $v \in T_pY$, the differential
\begin{align*}
d\iota_p: T_pY \to T_{\iota(p)}X
\end{align*}
is injective and complex-linear. Hence $d\iota_p(v) \neq 0$, and the positivity of the Kähler form $\omega$ gives
\begin{align*}
\eta_p(v,J_Yv)
=
\omega_{\iota(p)}(d\iota_p(v),J_Xd\iota_p(v))
>
0.
\end{align*}
Thus $\eta$ is a Kähler form on $Y$.
[guided]
We first move the problem entirely onto $Y$. Let $\iota:Y\hookrightarrow X$ denote the inclusion map, and let $J_X:TX\to TX$ and $J_Y:TY\to TY$ denote the complex structures on $X$ and $Y$, respectively. Define the pulled-back two-form
\begin{align*}
\eta := \iota^*\omega \in \Omega^2(Y).
\end{align*}
Because exterior differentiation commutes with pullback, and because $\omega$ is closed on $X$, we have
\begin{align*}
d\eta = d(\iota^*\omega) = \iota^*(d\omega) = 0.
\end{align*}
The inclusion $\iota: Y \hookrightarrow X$ is holomorphic, so it preserves complex tangent directions. Therefore the pullback of the real $(1,1)$-form $\omega$ is again a real $(1,1)$-form on $Y$.
It remains to check positivity. Fix $p \in Y$ and a nonzero vector $v \in T_pY$. The differential of the inclusion is the injective complex-[linear map](/page/Linear%20Map)
\begin{align*}
d\iota_p: T_pY \to T_{\iota(p)}X.
\end{align*}
Since $v \neq 0$, injectivity gives $d\iota_p(v) \neq 0$. Since $d\iota_p$ is complex-linear, it intertwines the complex structures:
\begin{align*}
d\iota_p(J_Yv)=J_Xd\iota_p(v).
\end{align*}
Therefore
\begin{align*}
\eta_p(v,J_Yv)
=
\omega_{\iota(p)}(d\iota_p(v),d\iota_p(J_Yv))
=
\omega_{\iota(p)}(d\iota_p(v),J_Xd\iota_p(v))
>
0,
\end{align*}
where the final inequality is exactly the positivity condition for the Kähler form $\omega$ on $X$. Hence $\eta$ is a Kähler form on $Y$.
[/guided]
[/step]
[step:Identify the top wedge as a positive volume form]
Fix $p \in Y$. Choose holomorphic coordinates
\begin{align*}
z_j = x_j + i y_j, \qquad 1 \leq j \leq m,
\end{align*}
on a coordinate neighbourhood $U \subset Y$ of $p$. At $p$, define the real [bilinear form](/page/Bilinear%20Form) $g_p:T_pY\times T_pY\to\mathbb{R}$ by
\begin{align*}
g_p(v,w):=\eta_p(v,J_Yw).
\end{align*}
The positivity of $\eta$ says that $g_p$ is a positive Hermitian form on the complex [vector space](/page/Vector%20Space) $T_pY$. Diagonalising this Hermitian form by a unitary change of coordinates gives [real numbers](/page/Real%20Numbers) $\lambda_1,\dots,\lambda_m > 0$ such that
\begin{align*}
\eta_p
=
\sum_{j=1}^m \lambda_j\, dx_j \wedge dy_j
\end{align*}
with respect to the complex orientation on $T_pY$. Therefore
\begin{align*}
\eta_p^m
=
m!\lambda_1\cdots\lambda_m\,
dx_1 \wedge dy_1 \wedge \cdots \wedge dx_m \wedge dy_m.
\end{align*}
Since each $\lambda_j>0$, the top-degree form $\eta^m$ is pointwise a positive multiple of the complex orientation volume form on $Y$.
[guided]
The purpose of this step is to turn positivity of a $(1,1)$-form into positivity of a top-degree form. Fix a point $p \in Y$ and choose holomorphic coordinates on a neighbourhood $U \subset Y$:
\begin{align*}
z_j = x_j + i y_j, \qquad 1 \leq j \leq m.
\end{align*}
The real coordinates
\begin{align*}
(x_1,y_1,\dots,x_m,y_m)
\end{align*}
determine the complex orientation on $Y$.
At the point $p$, define the real bilinear form $g_p:T_pY\times T_pY\to\mathbb{R}$ by
\begin{align*}
g_p(v,w):=\eta_p(v,J_Yw).
\end{align*}
Because $\eta$ is a positive real $(1,1)$-form, $g_p$ is a positive Hermitian form on the complex vector space $T_pY$. A positive Hermitian form can be diagonalised by a unitary change of complex coordinates. Thus, after such a coordinate change at $p$, there are real numbers $\lambda_1,\dots,\lambda_m>0$ such that
\begin{align*}
\eta_p
=
\sum_{j=1}^m \lambda_j\, dx_j \wedge dy_j.
\end{align*}
Now compute the $m$-fold wedge product. Terms with a repeated factor vanish because a differential form wedges with itself to zero. Hence only the terms using each pair $dx_j \wedge dy_j$ exactly once remain, and there are $m!$ such permutations. Therefore
\begin{align*}
\eta_p^m
=
m!\lambda_1\cdots\lambda_m\,
dx_1 \wedge dy_1 \wedge \cdots \wedge dx_m \wedge dy_m.
\end{align*}
The coefficient $m!\lambda_1\cdots\lambda_m$ is strictly positive. Hence $\eta^m$ is a positive smooth top-degree form with respect to the complex orientation on $Y$.
[/guided]
[/step]
[step:Integrate the positive top-degree form over the compact manifold]
Let $\mu_{\eta^m}$ denote the positive Borel measure on $Y$ induced by the positive smooth top-degree form $\eta^m$ and the complex orientation. Since $Y$ is compact and nonempty, this measure has strictly positive total mass. Indeed, choose any point $p \in Y$. By positivity and smoothness, there is a coordinate neighbourhood $U \subset Y$ of $p$ whose closure is compact in a coordinate chart and such that $\eta^m$ is a positive multiple of the coordinate orientation form throughout $U$. Therefore
\begin{align*}
\int_U 1\,d\mu_{\eta^m} > 0.
\end{align*}
Since $\eta^m$ is nonnegative with respect to the complex orientation on all of $Y$, enlarging the domain from $U$ to $Y$ gives
\begin{align*}
\int_Y 1\,d\mu_{\eta^m} \geq \int_U 1\,d\mu_{\eta^m} > 0.
\end{align*}
Recalling that $\eta=\iota^*\omega$, this proves
\begin{align*}
\int_Y 1\,d\mu_{(\iota^*\omega)^m} > 0,
\end{align*}
where $\mu_{(\iota^*\omega)^m}$ denotes the positive Borel measure induced by $(\iota^*\omega)^m$.
[guided]
We now turn pointwise positivity into strict positivity of the integral. Let $\mu_{\eta^m}$ be the positive Borel measure on $Y$ induced by the positive smooth top-degree form $\eta^m$ and the complex orientation. This means that, in any positively oriented coordinate chart, $\eta^m$ is written as a positive smooth density times the coordinate volume form, and $\mu_{\eta^m}$ is the corresponding measure.
Since $Y$ is nonempty, choose $p \in Y$. The preceding step shows that $\eta_p^m$ is a positive multiple of the complex orientation form at $p$. By smoothness, after shrinking to a coordinate neighbourhood $U \subset Y$ of $p$ whose closure is compact in the coordinate chart, the same coefficient remains positive throughout $U$. Hence
\begin{align*}
\int_U 1\,d\mu_{\eta^m} > 0.
\end{align*}
The form $\eta^m$ is nonnegative with respect to the complex orientation on all of $Y$, so enlarging the domain of integration from $U$ to $Y$ gives
\begin{align*}
\int_Y 1\,d\mu_{\eta^m} \geq \int_U 1\,d\mu_{\eta^m} > 0.
\end{align*}
Since $\eta=\iota^*\omega$, this is exactly
\begin{align*}
\int_Y 1\,d\mu_{(\iota^*\omega)^m} > 0,
\end{align*}
where $\mu_{(\iota^*\omega)^m}$ denotes the positive Borel measure induced by $(\iota^*\omega)^m$.
[/guided]
[/step]
[step:Interpret the integral as the cohomological pairing]
Let $H^{2m}(X;\mathbb{R})$ denote the degree-$2m$ de Rham cohomology group of $X$ with real coefficients. Let $[\omega]^m\in H^{2m}(X;\mathbb{R})$ denote the cup product of $m$ copies of the de Rham class $[\omega]\in H^2(X;\mathbb{R})$. Let $[Y]$ denote the fundamental homology class determined by the complex orientation of the compact real $2m$-manifold $Y$.
Because $\omega$ is closed, $\omega^m$ is closed. The pullback satisfies
\begin{align*}
\iota^*(\omega^m) = (\iota^*\omega)^m.
\end{align*}
Let $\mu_{\iota^*(\omega^m)}$ denote the positive Borel measure induced by the top-degree form $\iota^*(\omega^m)$ on the complex-oriented manifold $Y$. By the definition of the de Rham pairing between $H^{2m}(X;\mathbb{R})$ and the homology class represented by the oriented submanifold $Y$, one has
\begin{align*}
\langle [\omega]^m,[Y]\rangle
=
\int_Y 1\,d\mu_{\iota^*(\omega^m)}
=
\int_Y 1\,d\mu_{(\iota^*\omega)^m}.
\end{align*}
The preceding step shows that this number is strictly positive. Hence, for every positive-dimensional compact complex submanifold $Y \subset X$, the top-degree power $[\omega]^m$ pairs positively with its fundamental class.
[guided]
We now identify the positive integral with the cohomological pairing. Let $H^{2m}(X;\mathbb{R})$ denote the degree-$2m$ de Rham cohomology group of $X$ with real coefficients. Let $[\omega]^m\in H^{2m}(X;\mathbb{R})$ denote the cup product of $m$ copies of the de Rham class $[\omega]\in H^2(X;\mathbb{R})$. Let $[Y]$ denote the fundamental homology class determined by the complex orientation of the compact real $2m$-manifold $Y$.
The form $\omega$ is closed because it is Kähler. Therefore its wedge power $\omega^m$ is closed, since exterior differentiation satisfies the graded Leibniz rule. Pullback commutes with wedge products, so
\begin{align*}
\iota^*(\omega^m) = (\iota^*\omega)^m.
\end{align*}
Let $\mu_{\iota^*(\omega^m)}$ be the positive Borel measure induced by the top-degree form $\iota^*(\omega^m)$ on the complex-oriented manifold $Y$. The de Rham pairing of a degree-$2m$ cohomology class with the fundamental class of an oriented compact real $2m$-manifold represented by $Y$ is computed by integrating the pulled-back top-degree representative over $Y$. Hence
\begin{align*}
\langle [\omega]^m,[Y]\rangle
=
\int_Y 1\,d\mu_{\iota^*(\omega^m)}
=
\int_Y 1\,d\mu_{(\iota^*\omega)^m}.
\end{align*}
The integration step proved that the final quantity is strictly positive. Therefore the top-degree power $[\omega]^m$ pairs positively with the fundamental class $[Y]$.
[/guided]
[/step]
