[proofplan]
Choose an arbitrary Kähler class $\alpha$ and represent it by a Kähler form $\omega$. Since $c_1(X)=0$ in real cohomology, the zero real closed $(1,1)$-form represents $2\pi c_1(X)$, so it is an admissible prescribed Ricci form in Yau's theorem. Applying [citetheorem:9135] with $\rho=0$ gives a unique Kähler form in the class $[\omega]=\alpha$ whose Ricci form is zero.
[/proofplan]
[step:Choose a Kähler representative of the prescribed class]
Let $\alpha \in H^2(X;\mathbb R)$ be a Kähler cohomology class. By definition of a Kähler class, there exists a Kähler form $\omega$ on $X$ such that
\begin{align*}
[\omega]=\alpha
\end{align*}
in $H^2(X;\mathbb R)$.
[guided]
We begin with the object quantified in the theorem: a Kähler cohomology class $\alpha \in H^2(X;\mathbb R)$. The phrase "Kähler class" means precisely that $\alpha$ is represented by at least one Kähler form. Therefore we may choose a Kähler form $\omega$ on $X$ satisfying
\begin{align*}
[\omega]=\alpha.
\end{align*}
This choice is only a starting representative. The goal is not to prove that this particular $\omega$ is Ricci-flat, but to use it to specify the fixed cohomology class in which Yau's theorem will produce the desired representative.
[/guided]
[/step]
[step:Verify that the zero form has the required Ricci cohomology class]
Let $\rho$ denote the real closed $(1,1)$-form on $X$ given by $\rho=0$. Since $c_1(X)=0$ in $H^2(X;\mathbb R)$, its real multiple $2\pi c_1(X)$ is also zero. Hence $\rho$ represents the class $2\pi c_1(X)$:
\begin{align*}
[\rho]=0=2\pi c_1(X)
\end{align*}
in $H^2(X;\mathbb R)$.
[/step]
[step:Apply Yau's prescribed Ricci form theorem in the chosen class]
The hypotheses of [citetheorem:9135] are satisfied: $X$ is compact Kähler, $\omega$ is a Kähler form, and $\rho=0$ is a closed real $(1,1)$-form representing $2\pi c_1(X)$. Therefore there exists a unique Kähler form $\omega_\alpha$ such that
\begin{align*}
\omega_\alpha \in [\omega]
\end{align*}
and
\begin{align*}
\operatorname{Ric}(\omega_\alpha)=\rho.
\end{align*}
Since $\rho=0$, this gives
\begin{align*}
\operatorname{Ric}(\omega_\alpha)=0.
\end{align*}
[guided]
Now we invoke the exact form of Yau's theorem needed here, namely [citetheorem:9135]. That theorem applies to a compact Kähler manifold $X$, a Kähler form $\omega$, and a closed real $(1,1)$-form $\rho$ representing $2\pi c_1(X)$.
We verify these hypotheses one by one. The manifold $X$ is compact Kähler by assumption. The form $\omega$ is Kähler by the choice made in the first step. The form $\rho=0$ is a real closed $(1,1)$-form, and the previous step proved that it represents $2\pi c_1(X)$ because $c_1(X)=0$ in real cohomology. Hence all hypotheses of [citetheorem:9135] are satisfied.
The conclusion of Yau's theorem gives a unique Kähler form $\omega_\alpha$ in the cohomology class of $\omega$ such that
\begin{align*}
\operatorname{Ric}(\omega_\alpha)=\rho.
\end{align*}
Since $\rho$ is the zero form, this becomes
\begin{align*}
\operatorname{Ric}(\omega_\alpha)=0.
\end{align*}
This is the point where the vanishing of $c_1(X)$ is used: it makes the zero form a valid prescribed Ricci form. No trivialization of the canonical bundle is being assumed or needed.
[/guided]
[/step]
[step:Identify the resulting form with the original Kähler class and record uniqueness]
Because $\omega_\alpha \in [\omega]$ and $[\omega]=\alpha$, we have
\begin{align*}
[\omega_\alpha]=\alpha.
\end{align*}
Thus $\omega_\alpha$ is a Kähler form in the prescribed Kähler class with zero Ricci form.
If $\eta$ is another Kähler form satisfying $[\eta]=\alpha$ and $\operatorname{Ric}(\eta)=0$, then $\eta \in [\omega]$ and $\operatorname{Ric}(\eta)=\rho$. The uniqueness assertion in [citetheorem:9135], applied in the fixed class $[\omega]$, gives $\eta=\omega_\alpha$. Therefore the Ricci-flat Kähler form in the class $\alpha$ is unique.
[/step]
