The resultant detects finitely many common projective points

Let $K=\mathrm{Frac}(R)$ and regard $F,G\in K[x_2]$. They have no common factor of positive $x_2$-degree in $K[x_2]$: if a polynomial $H\in K[x_2]$ of positive degree divided both, then clearing denominators and applying Gauss's lemma to the primitive parts produces a nonconstant common divisor of $F$ and $G$ in $R[x_2]=k[x_0,x_1,x_2]$, contradicting the hypothesis.

For each $(a,b)\mathrm{\ne }(0,0)$ the specialised polynomials $F(a,b,x_2),G(a,b,x_2)\in k[x_2]$ have exact degrees $d$ and $e$, and ${\mathrm{Res}}_{x_2}(F,G)(a,b)=0$ if and only if they have a common root in $k$: the specialisation rule identifies the value with ${\mathrm{Res}}_{d,e}(F(a,b,x_2),G(a,b,x_2))$, and for binary forms of nominated positive degrees over the algebraically closed field $k$ the resultant vanishes exactly when a common zero in ${\mathbf{P}}_k^1$ exists. The point $[1:0]$ cannot be a common zero, since both specialised top coefficients are nonzero by hypothesis; thus the projective common zero is on the chart $[t:1]$ and is exactly a common finite root.

${\mathrm{Res}}_{x_2}(F,G)$ is a nonzero form of degree $de$. If it were the zero polynomial, then over the fraction field $K$ the Sylvester matrix would be singular, so there would be $A,B\in K[x_2]$, not both zero, with $\mathrm{deg}B\le d-1$, $\mathrm{deg}A\le e-1$ and $AF+BG=0$; then $F\mid BG$ in $K[x_2]$, and since $F$ and $G$ are coprime there by step 1.1 and $\mathrm{deg}B<d=\mathrm{deg}F$ (the top coefficient of $F$ in $x_2$ is a nonzero constant), this is impossible. Hence the determinant is nonzero, and the determinant-weight calculation in Resultant of two plane forms, viewed in one variable gives its total degree $de$ in $x_0,x_1$.

Let $Z$ be the zero set of ${\mathrm{Res}}_{x_2}(F,G)$ in ${\mathbf{P}}^1$. By step 1.2, for $(a,b)\mathrm{\ne }(0,0)$ the point $[a:b]$ lies in $Z$ exactly when $[a:b:c]\in C\cap D$ for some $c\in k$; since $[0:0:1]\mathrm{\notin }C\cup D$, every point of $C\cap D$ has $(a,b)\mathrm{\ne }(0,0)$, so the projection $\pi :C\cap D\to {\mathbf{P}}^1$ has image exactly $Z$. By step 2.1 the resultant is a nonzero form of degree $de$, so $Z$ is finite, of at most $de$ points: after possibly renaming the variables its dehomogenisation $\mathrm{Res}(1,t)$ is a nonzero polynomial of degree at most $de$, whose roots give the points of $Z$ with first coordinate nonzero, If the remaining point $[0:1]$ is a zero, the coefficient of $x_1^{de}$ vanishes, so the dehomogenisation has degree at most $de-1$; hence including that point still gives at most $de$ zeros. For $[a:b]\in Z$ the fibre consists of common roots of $F(a,b,x_2)$ and $G(a,b,x_2)$, hence of roots of the nonzero polynomial $F(a,b,x_2)$ of degree $d$ and also of $G(a,b,x_2)$ of degree $e$, so it has at most $\min (d,e)$ points. Therefore $C\cap D$ is finite with the asserted fibre bound, and its projections are exactly the zeros of the nonzero resultant.