[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*}[/step]

[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: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]