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!

For finite $I$ this is a consequence of Helly's Theorem. 
Without loss of generality your given lines are parallel to the $y$-axis, i.e., they have equations $x=x_i$ for $i\in I$. The segments have then the form 
$$s_i=\{(x_i,y)\ :\ y_i'\leqslant y\leqslant y_i''\}$$
The set of lines $y=ax+b$ intersecting segment $s_i$ is parameterized by the set
$$X_i=\{(a,b)\in\mathbb{R}^2\ :\ y_i'\leqslant ax_i+b\leqslant y_i''\}.$$
By assumption, the intersection of any three of the sets is non-empty, 
$$X_i\cap X_j\cap X_k\neq\emptyset\qquad\text{for all }i,j,k\in I.$$
Therefore, by Helly's theorem, the intersection of all the sets $X_i$ is nonempty,
$$\bigcap_{i\in I}X_i\neq\emptyset.$$
Any point $(a,b)$ in this intersection corresponds to a line intersecting all segments.
