[proofplan]
Choose a holomorphic line bundle isomorphism $\Phi:\mathcal O_X\to K_X$ and let $\Omega$ be the image of the constant section $1$ under $\Phi$. This gives a nowhere-vanishing holomorphic frame of $K_X$, so we can define a Hermitian metric on $K_X$ for which $\Omega$ has constant norm $1$. In local holomorphic coordinates the corresponding curvature form is zero, hence the real first Chern class of $K_X$ vanishes. Finally, since $K_X=\det T_X^*$, the determinant-dual relation for first Chern classes gives $c_1(K_X)=-c_1(T_X)$, so $c_1(X)=0$.
[/proofplan]
[step:Use the holomorphic line bundle isomorphism to produce a global frame of $K_X$]
Let
\begin{align*}
\Phi:\mathcal O_X \longrightarrow K_X
\end{align*}
be a holomorphic line bundle isomorphism. Let
\begin{align*}
1_X:X \longrightarrow \mathcal O_X
\end{align*}
denote the constant holomorphic section with value $1$ in every fibre of the holomorphically product line bundle. Define
\begin{align*}
\Omega:X \longrightarrow K_X
\end{align*}
by
\begin{align*}
\Omega(x)=\Phi_x(1_X(x))
\end{align*}
for every $x\in X$, where $\Phi_x:(\mathcal O_X)_x\to (K_X)_x$ is the induced fibre map.
Since $\Phi$ is holomorphic and $1_X$ is holomorphic, $\Omega$ is a holomorphic section of $K_X$. Since each $\Phi_x$ is a linear isomorphism and $1_X(x)\neq 0$ in $(\mathcal O_X)_x$, we have $\Omega(x)\neq 0$ for every $x\in X$. Thus $\Omega$ is a global nowhere-vanishing holomorphic frame of $K_X$.
[guided]
The hypothesis $K_X\cong\mathcal O_X$ means that there is a holomorphic line bundle isomorphism
\begin{align*}
\Phi:\mathcal O_X \longrightarrow K_X.
\end{align*}
The line bundle $\mathcal O_X$ has a distinguished global section
\begin{align*}
1_X:X \longrightarrow \mathcal O_X,
\end{align*}
namely the section whose value in each fibre is the scalar $1$. We transport this section through the isomorphism $\Phi$ and define
\begin{align*}
\Omega:X \longrightarrow K_X
\end{align*}
by
\begin{align*}
\Omega(x)=\Phi_x(1_X(x))
\end{align*}
for every $x\in X$.
This section is holomorphic because it is the composition of the holomorphic section $1_X$ with the holomorphic bundle map $\Phi$. It is nowhere vanishing because, for each $x\in X$, the fibre map
\begin{align*}
\Phi_x:(\mathcal O_X)_x\longrightarrow (K_X)_x
\end{align*}
is a linear isomorphism, and a linear isomorphism sends a nonzero vector to a nonzero vector. Since $1_X(x)\neq 0$, we get $\Omega(x)\neq 0$. Therefore $\Omega$ is a global holomorphic frame of the line bundle $K_X$.
[/guided]
[/step]
[step:Build a flat Hermitian metric on $K_X$ from the global frame]
Define a Hermitian metric $h$ on $K_X$ by requiring
\begin{align*}
h_x(\Omega(x),\Omega(x))=1
\end{align*}
for every $x\in X$, and extending fibrewise by Hermitian linearity. Explicitly, if $v\in (K_X)_x$, then there is a unique scalar $a\in\mathbb C$ such that
\begin{align*}
v=a\,\Omega(x),
\end{align*}
and we set
\begin{align*}
h_x(v,v)=|a|^2.
\end{align*}
Let $(U,z)$ be a holomorphic coordinate chart on $X$, where
\begin{align*}
z:U \longrightarrow z(U)\subseteq\mathbb C^n
\end{align*}
has coordinate functions $z_1,\dots,z_n$. The standard local holomorphic frame of $K_X$ over $U$ is
\begin{align*}
\kappa_U=dz_1\wedge \cdots \wedge dz_n.
\end{align*}
Since $\Omega$ is nowhere vanishing, there is a nowhere-vanishing [holomorphic function](/page/Holomorphic%20Function)
\begin{align*}
f_U:U \longrightarrow \mathbb C^*
\end{align*}
such that
\begin{align*}
\Omega=f_U\kappa_U.
\end{align*}
The metric coefficient of $h$ in the frame $\kappa_U$ is
\begin{align*}
h_U=h(\kappa_U,\kappa_U)=|f_U|^{-2}.
\end{align*}
The Chern curvature of a Hermitian holomorphic line bundle in a local holomorphic frame with metric coefficient $h_U$ is
\begin{align*}
\Theta_h=\bar\partial\partial \log h_U.
\end{align*}
Here
\begin{align*}
\log h_U=-\log |f_U|^2.
\end{align*}
Because $f_U$ is holomorphic and nowhere vanishing, locally $f_U=e^{g_U}$ for a holomorphic function
\begin{align*}
g_U:V \longrightarrow \mathbb C
\end{align*}
on each sufficiently small [open set](/page/Open%20Set) $V\subset U$. On such $V$,
\begin{align*}
\log |f_U|^2=g_U+\overline{g_U}.
\end{align*}
Therefore
\begin{align*}
\bar\partial\partial\log |f_U|^2=0,
\end{align*}
and hence $\Theta_h=0$ on $V$. Since these open sets cover $X$, the curvature form of $h$ is identically zero on $X$.
[/step]
[step:Conclude that the first Chern class of $K_X$ vanishes]
By the Chern-Weil definition of the real first Chern class of a Hermitian holomorphic line bundle, the closed real $(1,1)$-form
\begin{align*}
\frac{i}{2\pi}\Theta_h
\end{align*}
represents $c_1(K_X)$ in $H^2(X;\mathbb R)$. Since $\Theta_h=0$, this representative is the zero form. Hence
\begin{align*}
c_1(K_X)=0
\end{align*}
in $H^2(X;\mathbb R)$.
[/step]
[step:Use $K_X=\det T_X^*$ to identify $c_1(K_X)$ with $-c_1(X)$]
By definition of the canonical bundle,
\begin{align*}
K_X=\Lambda^n T_X^*=\det T_X^*.
\end{align*}
For a holomorphic vector bundle $E\to X$, the first Chern class of the determinant line bundle is the first Chern class of $E$, and dualization changes the sign of the first Chern class:
\begin{align*}
c_1(\det E)=c_1(E),
\end{align*}
and
\begin{align*}
c_1(E^*)=-c_1(E).
\end{align*}
Applying these identities to $E=T_X$ gives
\begin{align*}
c_1(K_X)=c_1(\det T_X^*)=c_1(T_X^*)=-c_1(T_X).
\end{align*}
Since $c_1(X):=c_1(T_X)$ and the previous step proved $c_1(K_X)=0$, we obtain
\begin{align*}
0=c_1(K_X)=-c_1(X).
\end{align*}
Therefore
\begin{align*}
c_1(X)=0
\end{align*}
in $H^2(X;\mathbb R)$, as required.
[/step]
