Suppose $C\subseteq \mathbb{R}^3$ is an irreducible real-algebraic curve of degree $k$ and $P\subset \mathbb{R}^3$ is a plane. Suppose that $C$ intersects $P$ finitely many times. What is the best way to upper bound the number of points of intersection of $C$ with $P$? Is it $k+1$?
Here's one method to provide an upper bound. Find a projection that collapses $P$ to a line. Use Tarski-Seidenberg theorem to see that $C$ projects to a semi-algebraic set of degree at most $c_k$. Now use Bezout's theorem to determine that $\pi(P)$ has at most $c_k+1$ points of intersection with $\pi(C).$ Show that the projection $\pi$ maps together at most $d_k$ points from $C$ (this last step is a little sketchy), so that $C$ and $P$ have at most $c_k d_k$ points of intersection.
Is there a way to use Bezout's theorem more directly? Or a better method?