[proofplan]
We compute the Hopf curvature in the affine chart $z_0\ne 0$ using the normalized local section of $S^3\to\mathbb{CP}^1$. Pulling the connection form back to this chart gives an explicit $i\mathbb R$-valued one-form, and differentiating it produces the local representative of the descended curvature. Comparing this expression with the normalized Fubini-Study form fixes the sign. The first Chern class conclusion then follows from the Chern-Weil representative for a Hermitian line bundle and from the normalization that $[\omega_{FS}]$ is the positive generator.
[/proofplan]
[step:Pull the Hopf connection back along the standard affine section]
Let
\begin{align*}
s:U_0&\to S^3
\end{align*}
\begin{align*}
[w]&\mapsto \frac{(1,w)}{(1+|w|^2)^{1/2}}
\end{align*}
denote the standard smooth local section, where $[w]$ denotes the point $[1:w]\in U_0$. Write
\begin{align*}
r:\mathbb C&\to \mathbb R_{>0}
\end{align*}
\begin{align*}
w&\mapsto (1+|w|^2)^{-1/2}.
\end{align*}
Then, along $s$, the coordinate functions on $S^3$ are $z_0=r$ and $z_1=wr$. Since $d(|w|^2)=\bar w\,dw+w\,d\bar w$, we have
\begin{align*}
dr=-\frac{1}{2}(1+|w|^2)^{-3/2}(\bar w\,dw+w\,d\bar w).
\end{align*}
Therefore
\begin{align*}
s^*\alpha=r\,dr+\bar w r(r\,dw+w\,dr).
\end{align*}
Combining the two $dr$ terms gives
\begin{align*}
s^*\alpha=(1+|w|^2)r\,dr+r^2\bar w\,dw.
\end{align*}
Substituting $r^2=(1+|w|^2)^{-1}$ and the formula for $dr$ yields
\begin{align*}
s^*\alpha=-\frac{1}{2}\frac{\bar w\,dw+w\,d\bar w}{1+|w|^2}+\frac{\bar w\,dw}{1+|w|^2}.
\end{align*}
Thus
\begin{align*}
s^*\alpha=\frac{\bar w\,dw-w\,d\bar w}{2(1+|w|^2)}.
\end{align*}
[guided]
The purpose of this step is to turn the global connection form into a concrete one-form in the single complex coordinate $w$. The section
\begin{align*}
s:U_0&\to S^3
\end{align*}
\begin{align*}
[w]&\mapsto \frac{(1,w)}{(1+|w|^2)^{1/2}}
\end{align*}
is well-defined because every point in $U_0$ has a unique representative of the form $[1:w]$, and the vector $(1,w)/(1+|w|^2)^{1/2}$ has norm $1$ in $\mathbb C^2$.
Define
\begin{align*}
r:\mathbb C&\to \mathbb R_{>0}
\end{align*}
\begin{align*}
w&\mapsto (1+|w|^2)^{-1/2}.
\end{align*}
Along the section, $z_0=r$ and $z_1=wr$. The connection form is
\begin{align*}
\alpha=\bar z_0\,dz_0+\bar z_1\,dz_1.
\end{align*}
Since $r$ is real-valued, $\bar z_0=r$, and since $z_1=wr$, we have $\bar z_1=\bar w r$ and $dz_1=r\,dw+w\,dr$. Hence
\begin{align*}
s^*\alpha=r\,dr+\bar w r(r\,dw+w\,dr).
\end{align*}
The terms containing $dr$ combine because $r\,dr+\bar w w r\,dr=(1+|w|^2)r\,dr$. Therefore
\begin{align*}
s^*\alpha=(1+|w|^2)r\,dr+r^2\bar w\,dw.
\end{align*}
Now compute $dr$ from the definition of $r$. Since
\begin{align*}
d(|w|^2)=d(w\bar w)=\bar w\,dw+w\,d\bar w,
\end{align*}
the ordinary chain rule gives
\begin{align*}
dr=-\frac{1}{2}(1+|w|^2)^{-3/2}(\bar w\,dw+w\,d\bar w).
\end{align*}
Substituting this and $r^2=(1+|w|^2)^{-1}$ into the expression for $s^*\alpha$ gives
\begin{align*}
s^*\alpha=-\frac{1}{2}\frac{\bar w\,dw+w\,d\bar w}{1+|w|^2}+\frac{\bar w\,dw}{1+|w|^2}.
\end{align*}
Collecting the $dw$ and $d\bar w$ terms yields
\begin{align*}
s^*\alpha=\frac{\bar w\,dw-w\,d\bar w}{2(1+|w|^2)}.
\end{align*}
This formula is $i\mathbb R$-valued, as required for a $U(1)$-connection, because complex conjugation changes the numerator to its negative.
[/guided]
[/step]
[step:Differentiate the local connection form to compute the descended curvature]
Let
\begin{align*}
\beta:=s^*\alpha\in \Omega^1(U_0;i\mathbb R).
\end{align*}
By the pullback compatibility of exterior differentiation,
\begin{align*}
s^*d\alpha=d\beta.
\end{align*}
Using
\begin{align*}
\beta=\frac{1}{2}(1+|w|^2)^{-1}(\bar w\,dw-w\,d\bar w),
\end{align*}
the product rule gives
\begin{align*}
d\beta=\frac{1}{2}d\bigl((1+|w|^2)^{-1}\bigr)\wedge(\bar w\,dw-w\,d\bar w)+\frac{1}{2}(1+|w|^2)^{-1}d(\bar w\,dw-w\,d\bar w).
\end{align*}
First,
\begin{align*}
d\bigl((1+|w|^2)^{-1}\bigr)=-(1+|w|^2)^{-2}(\bar w\,dw+w\,d\bar w).
\end{align*}
Second, since $d(dw)=d(d\bar w)=0$,
\begin{align*}
d(\bar w\,dw-w\,d\bar w)=d\bar w\wedge dw-dw\wedge d\bar w=2\,d\bar w\wedge dw.
\end{align*}
Substituting these identities and using $dw\wedge dw=d\bar w\wedge d\bar w=0$ gives
\begin{align*}
d\beta=-\frac{1}{2}(1+|w|^2)^{-2}\bigl(2|w|^2\,d\bar w\wedge dw\bigr)+(1+|w|^2)^{-1}d\bar w\wedge dw.
\end{align*}
Therefore
\begin{align*}
d\beta=\frac{d\bar w\wedge dw}{(1+|w|^2)^2}.
\end{align*}
Since $\pi\circ s=\operatorname{id}_{U_0}$ and $\pi^*\Omega=d\alpha$, we also have
\begin{align*}
\Omega\big|_{U_0}=s^*\pi^*\Omega=s^*d\alpha=d\beta=\frac{d\bar w\wedge dw}{(1+|w|^2)^2}.
\end{align*}
[/step]
[step:Compare the curvature form with the normalized Fubini-Study form]
Because $d\bar w\wedge dw=-dw\wedge d\bar w$, the preceding curvature formula gives
\begin{align*}
\Omega\big|_{U_0}=-\frac{dw\wedge d\bar w}{(1+|w|^2)^2}.
\end{align*}
Multiplying by $i/(2\pi)$ gives
\begin{align*}
\frac{i}{2\pi}\Omega\big|_{U_0}=-\frac{i}{2\pi}\frac{dw\wedge d\bar w}{(1+|w|^2)^2}.
\end{align*}
By the normalization of the Fubini-Study form in the theorem statement,
\begin{align*}
\omega_{FS}\big|_{U_0}=\frac{i}{2\pi}\frac{dw\wedge d\bar w}{(1+|w|^2)^2}.
\end{align*}
Hence
\begin{align*}
\frac{i}{2\pi}\Omega\big|_{U_0}=-\omega_{FS}\big|_{U_0}.
\end{align*}
The chart $U_0$ is dense in $\mathbb{CP}^1$, and both sides are smooth $2$-forms on $\mathbb{CP}^1$. Since smooth forms that agree on a dense open subset agree everywhere, we obtain
\begin{align*}
\frac{i}{2\pi}\Omega=-\omega_{FS}.
\end{align*}
[/step]
[step:Identify the Chern-Weil class of the tautological line bundle]
The tautological line bundle $\mathcal O(-1)\to \mathbb{CP}^1$ is the complex line bundle associated to the Hopf principal $U(1)$-bundle by the standard defining representation convention. The connection $\alpha$ induces the corresponding unitary connection on $\mathcal O(-1)$, and its curvature is the descended form $\Omega$. By [citetheorem:9800], the first Chern form of this unitary connection is
\begin{align*}
\frac{i}{2\pi}\Omega.
\end{align*}
Therefore the de Rham image of $c_1(\mathcal O(-1))$ is
\begin{align*}
\left[\frac{i}{2\pi}\Omega\right]=[-\omega_{FS}]=-[\omega_{FS}]
\end{align*}
in $H^2_{\mathrm{dR}}(\mathbb{CP}^1)$.
By the normalization in the theorem statement, $u\in H^2(\mathbb{CP}^1;\mathbb Z)$ is the positive generator whose image under the de Rham comparison map is $[\omega_{FS}]$. Hence the integral class with de Rham image $-[\omega_{FS}]$ is $-u$. Consequently
\begin{align*}
c_1(\mathcal O(-1))=-u.
\end{align*}
This proves both the curvature identity and the asserted first Chern class formula.
[/step]
