My question is a bit related to both the container method and shallow cell complexity.
Let's start with that the number of length $\ell$ paths (where $\ell$ denotes the number of *vertices* of the path!) in a planar graph on $n$ vertices is $O(n^{\lfloor\frac{\ell+1}2\rfloor})$ (where the hidden constant depends on $\ell$). This follows from that there are $O(n)$ choices for every second edge. It is also best possible as shown by blowing up every second vertex of a path of length $\ell$ to $n/\ell$ vertices. Now I'll state my question.

Is it true that in a planar graph on $n$ vertices one can select $O(n)$ ${\lceil\frac{\ell+1}2\rceil}$-tuples of vertices such that any path of length $\ell$ contains one of these ${\lceil\frac{\ell+1}2\rceil}$-tuples among its vertices?

Note that this would also imply that the number of length $\ell$ paths is $O(n\cdot n^{\ell-\lceil\frac{\ell+1}2\rceil})=O(n^{\lfloor\frac{\ell+1}2\rfloor})$, as there are $O(n^{\ell-\lceil\frac{\ell+1}2\rceil})$ ways one can select the remaining vertices of the path. Another, probably nicer way to state the question is if we let $k=\lceil\frac{\ell+1}2\rceil$.

Is it true that in a planar graph on $n$ vertices one can select $O(n)$ $k$-tuples of vertices such that any path of length $2k-2$ contains one of these $k$-tuples among its vertices?

This is trivial for $k=1,2$ as the number of vertices/edges is $O(n)$. That we cannot hope to hit all paths of length $2k-3$ by $k$-tuples is shown by the same example as above; blow up every second vertex of a path of length $2k-3$ to $n/k$ vertices.

I couldn't even prove my question for $k=3$, nor show that it would hold with some other function $f(k)$ instead of $2k-2$.