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A slightly different way to prove Francesco+auniket's result is as follows. First, given two different planes in the family, all $d$ points must lie in their intersection, which is a line $L$.
Second, every plane intersecting $C$ properly does so in $d$ points (counting multiplicities). So for every plane $H$ through $L$ intersecting $C$ properly, $H\cap C \subset L$. If there is any such plane, then general planes through $L$ intersect $C$ properly. Let $H_1, \dots, H_m$ be the (therefore finitely many) planes through $L$ which meet $C$ nonproperly. Then $C \subset \bigcup H_i$, each $H_i$ contains a component $C_i$ of $C$ of degree $d_i$ which goes through $d_i$ of the $d$ points. If there is no such plane on the other hand, then $L$ is a component of $C$.
Actually, this is projecting from $L$ rather than projecting from a point of $L$.
A slightly different way to prove Francesco+auniket's result is as follows. First, given two different planes in the family, all $d$ points must lie in their intersection, which is a line $L$.
Second, every plane intersecting $C$ properly does so in $d$ points (counting multiplicities). So for every plane $H$ through $L$ intersecting $C$ properly, $H\cap C \subset L$. If there is any such plane, then general planes through $L$ intersect $C$ properly. Let $H_1, \dots, H_m$ be the (therefore finitely many) planes through $L$ which meet $C$ nonproperly. Then $C \subset \bigcup H_i$, each $H_i$ contains a component $C_i$ of $C$ of degree $d_i$ which goes through $d_i$ of the $d$ points. If there is no such plane on the other hand, then $L$ is a component of $C$.