A quintic rational curve X in P3 is a projection of a normal rational quintic C in P5 from a line r. A 4-secant line in P3 corresponds to a 4-secant P3, H, to C in P5 containing r.
(*) So we must prove that, given any line r, there is a H (= P3) containing r and 4-secant to C.
Observe that 4 points p1,...,p4 on C span a P (and not less otherwise C would be degenerated). So we get a morphism f:Hilb4(C)--> Gr(3,5). Here Hilb4(C) is the projective scheme parametrizing 4-uples of points on C (actually zero-dimensional subschemes of length 4; we can think of it as P(S4) where S4 is the vector space of homogeneous polynomials of degree 4 in two variables (the roots give a 4-uple on P1 = C, and viceversa)). We have dim(Hilb4(C))=4; also dim(Gr(3,5))=8.
Now consider the incidence variety I contained in Gr(1,5)xGr(3,5), I = {(l,H)/l is contained in H}, with its natural projections: p:I --> Gr(1,5), q:I--> Gr(3,5).
We are interested in J:=f*(I), set theoretically J = {(l,H)/ l is contained in H and where H = <p1,...,p4> where pi are points on C}, i.e. J parametrizes lines in the P3 spanned by four points on C (i.e. 4-secant P3 to C). The fibers of J-->Hilb4(C) are isomorphic to Gr(1,3) which has dimension 4. It turns out that J is irreducible of dimension 8.
Now we have a natural morphism h: J --> Gr(1,5):(H=<p1,...,p4>,l) --> l and statement (*) is equivalent to show that h is surjective.
To show that h is surjective we argue as follows:
So it is enough to show that dim(h(J)) has dimension 8; for this it is enough to show that the generic fiber of h has dimension 0 ([Sha]??).
This in turn is equivalent to the existence of a quintic rational curve, X in P3, with a finite (>0) number of 4-secant lines.
Take a general divisor of multi-degree (7;4,3,3,2,2,2) on a smooth cubic surface S; this divisor is a smooth rational 5-ic curve, X. The exceptional divisor E1 corresponds to a 4-secant line to X. Moreover since every 4-secant to X is contained in S and since S contains only 27 lines, X has a finite number of 4-secant lines.
This concludes the proof of the existence of the 4-secant line.
Remark: The existence of a 4-secant line follows directly from a very general result of Gruson-Lazarsfeld-Peskine: "
Now we show that if h0(IC(2))=0, then C has a unique 4-secant line.
Assume R and D are two 4-secant lines to C. First R and D are necessarly disjoint (otherwise they would span a plane intersectiong C in too many points). Now observe that through any point of C there pass a trisecant to C (otherwise projectiong from that point we would get a smooth rational quartic plane curve: impossible); take L a trisecant to C. Arguing as before L has to be disjoint from R and D. There exists a smooth quadric Q containing L, R and D; since Q intersects C in at least 11 points it has to contain C; contradiction.