Setup & question
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Let $C \hookrightarrow \mathbb{P}^{g-1}$ be a general canonical curve of genus $g \ge 4$ and let $Y\_1,Y\_2 \subset \mathbb{P}^{g-1}$ be codimension 2 linear subspaces such that $Y\_i \cap C = \emptyset$ for $i=1,2$. Given any pair of points $(p,q) \in C^2$ we will denote by $L\_{p,q} \subset \mathbb{P}^{g-1}$ the bisecant line through $p$ and $q$ (or the tangent line if $p=q$).



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> Are there only finitely many pairs $(p,q) \in C^2$ such that $L\_{p,q}$ intersects both $Y\_1$ and $Y\_2$?
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A remark on the generality assumption
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Let me emphasize that $Y\_i$'s are not general except for the fact that they do not intersect $C$. In particular, $Y\_i$'s can be skew (i.e. they lie in a common hyperplane). However, $C$ is assumed to be general.


In fact the generality of $C$ is mandatory: Take $g=4$ and $C$ "special'' in the sense that the quadric surface containing $C$ is a cone. Denote the vertex of this cone by $x$. There is a pencil of lines through $x$ which are trisecants (these are the fibers of the projection from $x$). If we take $Y\_1$ and $Y\_2$ to be (non-skew) lines intersecting at $x$ then we have our counter-example.



Do only finitely many bisecants of a canonical curve intersect two distinct codimension 2 spaces simultaneously?
For every curve $C$ of genus $g\geq 5$ ~~that is neither hyperlliptic nor trigonal and that admits no morphism of degree greater than 1 to a curve of positive genus~~ that has general moduli, there exists no such pair $(Y\_1,Y\_2)$ of distinct linear spaces.


For every point $p$, for every codimension $2$ linear space $Y$ that does not contain $p$, the span of $p$ and $Y$ is a hyperplane. This hyperplane intersects $C$ in only finitely many points. Thus, there are only finitely many secant lines to $C$ that contain $p$ and intersect $Y$. Thus, for $Y\_1$ and $Y\_2$, there is no point $p$ such that infinitely many secant lines to $C$ contain $p$ and intersect both $Y\_1$ and $Y\_2$. So if there are infinitely many secant lines that intersect both $Y\_1$ and $Y\_2$, then for a general $p$ in $C$, there exists at least one such secant line that contains $p$.


For each of the two linear subspaces, consider the corresponding linear projection $$\pi\_j: \mathbb{P}^{g-1}\setminus Y\_j \to \Pi\_j, \ \ j=1,2,$$ where $\Pi\_j$ is a copy of $\mathbb{P}^1$. Consider the morphism $$\pi:C\to \Pi\_1\times \Pi\_2, \ \ \pi(p)=(\pi\_1(p),\pi\_2(p)).$$ By the previous paragraph, $\pi$ maps $C$ to its image as a morphism of degree $d\geq 2$. The image cycle in $\Pi\_1\times\Pi\_2$ has bidegree $(2g-2,2g-2)$. By hypothesis, $C$ admits no finite, surjective morphism to a curve of genus $h$ with $1\leq h\leq g-1$. Thus, the image curve must be a genus $0$ curve. Either the image curve is a smooth curve of bidegree $(1,1)$, or it is a singular curve.


If the image curve is a smooth curve, then it is the graph of an isomorphism from $\Pi\_1$ to $\Pi\_2$. Thus, the associated $2$-dimensional linear systems of canonical divisors on $C$ are equal, i.e., $Y\_1$ equals $Y\_2$ and the morphism $\pi$ factors through the diagonal. 


**Edit.** Since the curve has general moduli, the gonality of $C$ is $[g/2]+1$. Thus, the image curve either has bidegree $(2,2)$ or $(3,3)$. If the image curve has bidegree $(2,2)$, then projection away from a singular point defines an isomorphism to a plane conic. In other words, $C$ has a theta characteristic $L$ with $h^0(C,L) \geq 2$. This cannot happen if $C$ has general moduli by a Theorem of Montserrat Teixidor, cf. the following MathOverflow answer and reference: [Theta characteristics of genus$\geq3$ curve](https://mathoverflow.net/questions/210103/theta-characteristics-of-genus-geq3-curve)


MR0887499 (88e:14037)   
 
Teixidor i Bigas, Montserrat(E-BARU)   

Half-canonical series on algebraic curves.   

Trans. Amer. Math. Soc. 302 (1987), no. 1, 99–115.   
 
14H10   

<http://www.jstor.org/stable/2000899?origin=crossref&seq=1#page_scan_tab_contents>


I will have to think further about whether a general curve could have an invertible sheaf $L$ with $L^{\otimes 3}\cong \omega\_C$ and $h^0(C,L)\geq 2$, but it seems even less likely than having a theta characteristic with $h^0(C,L)\geq 2$. 


**Second edit.** Emre and Gregor Bruns finished the proof in the comments. Let me just summarize. The normalization $B$ of the image of $\pi$ is a smooth rational curve. For the associated morphism $$\widetilde{\pi}:C\to B,$$ $\widetilde{\pi}^\*\mathcal{O}\_B(1)$ is an invertible sheaf $\mathcal{L}$ on $C$ that has a basepoint free pencil of sections (defining the morphism to $B\cong \mathbb{P}^1$). The image of $B$ in $\Pi\_1\times \pi\_2$ is a curve of bidegree $(d\_1,d\_2)$ for integers $d\_1,d\_2\geq 0.$ Thus, $L^{\otimes d\_j}$ equals $\widetilde{\pi}^\*(\text{pr}\_j^\*\mathcal{O}\_{\Pi\_j}(1))$. This equals $\pi\_j^\*\mathcal{O}\_{\Pi\_j}(1)$, which in turn equals $\omega\_C$ since $\pi\_j$ is a linear projection of a canonical curve (and the center $Y\_j$ of the projection is disjoint from $C$). Since $\mathcal{L}^{\otimes d\_1}$ is isomorphic to $\mathcal{L}^{\otimes d\_2}$, also $d\_1$ equals $d\_2$; call this common integer $d$. Since the gonality of a general curve is $[g/2]+1$, also the degree $(2g-2)/d$ of $\mathcal{L}$ is at least $[g/2]+1$. Thus $d$ equals $1,$ $2,$ or $3.$ 


We want to prove that $d$ equals $1$, for then the pencil of sections of $\omega\_C$ coming from $\pi\_1$ and $\pi\_2$ are both the same: just the pencil of sections coming from $\widetilde{\pi}.$ This implies that the zero loci of those pencils are equal, i.e., $Y\_1$ equals $Y\_2$. So the strategy is to rule out $d=2$ and $d=3$ by contradiction. 


By the theorem of Montserrat Teixidor, for a general curve there is no invertible sheaf $\mathcal{L}$ on $C$ with both $h^0(C,\mathcal{L})\geq 2$ and with $\mathcal{L}^{\otimes 2}\cong \omega\_C$, i.e., $\mathcal{L}$ is a theta characteristic. This rules out $d=2$. 


Finally, as Gregor Bruns proves, also we cannot have an invertible sheaf $\mathcal{L}$ with both a basepoint free pencil of sections and with $\mathcal{L}^{\otimes 3}\cong \omega\_C$ for a curve of general moduli (hyperelliptic curves of genus $g = 3h+1$ do have such linear systems). 


Here is my interpretation of Gregor's argument, but I hope that Gregor will correct me if I am wrong. For any invertible sheaf $\mathcal{L}$ on $C$ with a basepoint free pencil of sections, $$\text{span}(s\_0,s\_1)\subset H^0(C,\mathcal{L}),$$ setting $\mathcal{F}=\omega\_C\otimes\mathcal{L}^\vee$, we can apply the Basepoint Free Pencil Trick (p. 126 of Arbarello-Cornalba-Griffiths-Harris) to see that the kernel of the following cup-product pairing equals $H^0(C,\omega\_C\otimes (\mathcal{L}^\vee)^{\otimes 2}) \cong H^1(C,\mathcal{L}^{\otimes 2})^\vee$: $$ \text{span}(s\_0,s\_1)\otimes H^0(C,\omega\_C\otimes\mathcal{L}^\vee) \to H^0(C,\omega\_C).$$ By the Gieseker-Petri Theorem (which implies smoothness of the parameter spaces of $\mathfrak{g}^r\_d$s), for a curve $C$ of general moduli, this map is injective. Thus $H^1(C,\mathcal{L}^{\otimes 2})$ is zero. On the other hand, if $\mathcal{L}^{\otimes 3}\cong \omega\_C$, then this group equals $H^1(C,\omega\_C\otimes \mathcal{L}^\vee) \cong H^0(C,\mathcal{L})^\vee$, by Serre duality. So the hypothesis that $\mathcal{L}$ has a basepoint free pencil of divisors leads to the contradictory conclusion that $\mathcal{L}$ has only the zero global section. This rules out $d=3$.


Altogether, this proves that for a general canonical curve $C$, there cannot exist infinitely many secant lines to $C$ that intersect both $Y\_1$ and $Y\_2$ unless $Y\_1$ equals $Y\_2$.


