Segments on a family of parallel lines Let $\{l_i:i\in I\}$ be a family of parallel lines on the plane $\mathbb{R}^2$. Suppose for each $i\in I$ there is a closed segment $s_i\subset l_i$. Moreover, for each triple $i_1,i_2,i_3$ there exists a line $l$ intersecting each of $s_{i_1},s_{i_2},s_{i_3}$. I would like to prove that there exists a line intersecting all segments $s_i$! I'd like to add that $I$ can be infinite.

It seems to be rather complicated even in the case of $4$ lines! Moreover, the statement is wrong if we replace segments by open intervals: there is no line intersecting each interval of the family $\{(0,\frac{1}{n})\times \{n\}: n\in \mathbb{N}\}$, but each triple of course can be intersected by some vertical line.
Can anybody please suggest something?

I was trying to prove some simple finite cases, but failed. I think it is important to neglect specificity of finiteness: the set of points of segments can even be dense!

Thomas Kalinowski has done the hard work in proving the result for finite $I$. To complete the argument, let $\mathscr{F}=\{X_i:i\in I\}$, where $I$ is arbitrary and $X_i$ is as in his answer. Each $X_i$ is a closed strip of constant width and slope $-x_i$. Fix distinct $i,j\in I$. Then $x_i\ne x_j$, so $X_i\cap X_j$ is a closed rhomboid $R$. Let $\mathscr{K}=\{X_k\cap R:k\in I\}$; $\mathscr{K}$ is then a centred family of compact sets in $\Bbb R^2$, so its intersection is non-empty, and any point $\langle a,b\rangle\in\bigcap\mathscr{K}$ corresponds to a line intersecting all segments.
