= Solution
Choose <homogeneous polynomials> $F,G$ defining the two <projective plane curves>. Their absence of a common <irreducible component> makes them coprime. For a point $O=[1:0:0]$ outside both <curves>, they have degrees $n,m$ in $X$ and nonzero constant leading coefficients. The <resultant>
$$
R(Y,Z)=\operatorname{Res}_X(F(X,Y,Z),G(X,Y,Z))
$$
is therefore nonzero. Indeed, a vanishing <resultant> over $k(Y,Z)$ would give a common factor there, and <Gauss's lemma> would give a common component of the original <curves>.
The <homogeneity> of $F,G$ makes $R$ a <homogeneous polynomial> of degree $nm$: scaling $(Y,Z)$ by $a$ scales their $X$-roots by $a$, and the product of the $nm$ differences of roots scales by $a^{nm}$. Every point in the <intersection> projects from $O$ to a zero of $R$ in the <projective line>. Each such projection line contains only finitely many points of either <curve>, since it passes through $O$ outside them. Thus the <intersection> is finite already.
We may now choose $O$ also outside the finitely many joining lines of distinct intersection points. Projection is then injective on the <intersection>. A nonzero degree-$nm$ <homogeneous polynomial> on the <projective line> has at most $nm$ distinct zeros, so the <resultant bound for intersections of plane curves> gives
$$
\boxed{|C\cap D|\le nm.}
$$
This derives the required cardinality bound without assuming the full intersection-multiplicity statement of <Bézout's theorem>.
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