A Geometric Combinatorial/Graph Theory Question I have a combinatorics problem that seems pretty general - I'd be surprised if the answer is not known.  Unfortunately, I can't seem to solve it.
The question concerns the following situation: Suppose we have a set $L$ composed of $k$ disjoint, parallel half-lines $L_i$ in $\mathbb{R}^3$ (drawn in blue below).  Suppose we have another collection of $n$ pairwise-disjoint paths $P_j$ in $\mathbb{R}^3$ (drawn in the other colors with $n = 4$) so that if $P_j = \gamma_j([0, \infty))$ then $\gamma_j(0) \in L$, $\gamma_j(x)$ escapes to infinity, and $P_j \cap L_i$ is infinite for each $i, j$.  Thus the $P_j$ are half-lines that bounce along the blue lines $L_i$ hitting each infinitely many times as they travel 'down' their length off to infinity.

The second image shows that in our example, by excising some initial sub-arcs of the $P_j$'s, and excising some terminal sub-arcs of the $L_i$'s, we can obtain $k$ pairwise-disjoint half-lines each of which 'begins' with an arc from some $L_i$.  The question is: For a given $L_k$, is there an $n(k)$ so large that if $P_j$, $1 \leq j \leq n(k)$ are as prescribed then we may pick some $k$ of them and, using the same excision procedure, obtain $k$ pairwise-disjoint half-lines each beginning with an arc from $L$?  I have been having some difficulty distilling the question into purely combinatorial terms.
This problem is coming from a problem in topology which I asked on MSE, and as exhibited there this result would be sufficient to prove a nice theorem in continuum theory:
https://math.stackexchange.com/questions/2344525/hairy-points-in-infinite-graphs-and-peano-continua
Thanks for any help!  I am useless at these sorts of things.
 A: I claim that $n(k)=k$. At first, I slightly reformulate the problem. 
We have a countable set $X$ of points, which is the intersection of a union of $k$ rays $r_1,\dots, r_k$, and a union of $k$ paths $p_1,\dots,p_k$. Each ray and each path is linearly ordered, and each of the sets $r_i\cap p_j\subset X$  is countable. Each (non-trivial) initial segment of a ray or a path contains only finitely many points from $X$. 
We may remove an initial part from any path, and we want to reach the following property: $k$ minima $x_1,\dots,x_k$ of the $k$ sets $X\cap r_i$ all belong to different paths. 
If they are not of different paths, say, green path is absent, we try to increase the number of different paths in the set $\{x_1,\dots,x_k\}$. If, say, points $x_1,x_2,x_7$ lie on a blue path, and $x_1$ appears on the blue path before $x_2$ and $x_7$, we remove the initial part of the blue path up to $x_1$ ($x_1$ is also removed). Note that the number of different colours in $\{x_1,\dots,x_k\}$ does not decrease (since $x_2$ and $x_7$ remain blue). Assume that it is not changed, that is, new $x_1$ belongs to the paths which already contains one of the points $x_2,\dots,x_k$. Proceed this way. Note that we may do only finitely many steps, because green minima on rays are "stop-points". So, after finitely many steps we increase the number of different paths in the set $\{x_1,\dots,x_k\}$, as desired.
