[guided]The point of the argument is to turn a positive form in the correct cohomology class into the curvature form of an actual Hermitian metric. We begin with an arbitrary smooth Hermitian metric $h_0$ on $L$. This gives a reference Chern form $c_1(L,h_0)$ representing $c_1^{BC}(L)$. The hypothesis gives another representative $\omega$ of the same Bott-Chern class, and the meaning of equality in Bott-Chern cohomology is exactly that their difference is globally of the form
\begin{align*}\omega-c_1(L,h_0)=\frac{i}{2\pi}\partial\bar\partial\psi\end{align*}
for some smooth real function $\psi:X\to\mathbb R$.
Now define $h=e^{-\psi}h_0$. This is again a smooth Hermitian metric because $e^{-\psi}$ is a smooth positive real-valued function on $X$. To compute its curvature, work on a holomorphic trivialising open set $U\subset X$ with frame $e:U\to L|_U$. Let $\varphi_{0,e}:U\to\mathbb R$ be the local weight of $h_0$, so
\begin{align*}|e|_{h_0}^2=\exp(-\varphi_{0,e}).\end{align*}
For the new metric $h$, we have
\begin{align*}|e|_h^2=e^{-\psi}|e|_{h_0}^2.\end{align*}
Substituting the expression for $|e|_{h_0}^2$ gives
\begin{align*}
|e|_h^2=\exp(-\psi)\exp(-\varphi_{0,e})=\exp(-(\varphi_{0,e}+\psi|_U)).
\end{align*}
Thus the local weight of $h$ is $\varphi_{0,e}+\psi|_U$.
Using the Chern-form formula for the local weight, we obtain
\begin{align*}
c_1(L,h)|_U=\frac{i}{2\pi}\partial\bar\partial(\varphi_{0,e}+\psi|_U).
\end{align*}
Linearity of $\partial\bar\partial$ gives
\begin{align*}
c_1(L,h)|_U=\frac{i}{2\pi}\partial\bar\partial\varphi_{0,e}+\frac{i}{2\pi}\partial\bar\partial\psi|_U.
\end{align*}
The first term is $c_1(L,h_0)|_U$, so
\begin{align*}
c_1(L,h)|_U=c_1(L,h_0)|_U+\frac{i}{2\pi}\partial\bar\partial\psi|_U.
\end{align*}
By the Bott-Chern potential identity, this is exactly $\omega|_U$. Since the computation is local and the forms agree on every holomorphic trivialisation, the global equality is
\begin{align*}c_1(L,h)=\omega.\end{align*}
Because $\omega$ was assumed positive, the metric $h$ has positive Chern form. Therefore $L$ is positive.[/guided]